4.3 CPT Interpretation Method
4.3.2 FE Simulations
FE Mesh
The FE mesh consists of 8000 4-node continuum elements with axial symmetry (Fig.
4.3). The upper and lower boundary of the specimen is fixed in vertical direction. At the outer boundary the corresponding cell pressure is applied. The CPT probe is modelled as a rigid tube inside the specimen along the symmetry axis. Underneath the ”cone”
the diameter of the tube is 1.8 mm, above the ”rod” has a diameter of 18 mm. When the probe is forced down, the soil is displaced by the cone. In order to avoid excessive distortions of the elements at the top surface, the cone is initially positioned inside the specimen. During the simulation the probe is displaced 0.25 cm down, which has been proven to be sufficient in order to achieve a kind of limit state (Fig. 4.4).
90 cm
30 cm
Figure 4.3: Finite element mesh for the simulation of CPT calibration chamber tests.
Finite sliding is allowed at the interface between soil and probe, but no separation. The contact is modelled with zero friction. This is due to the fact that friction is only desired at the cone (and when required at the sleeve) but not along the rod, which in reality
4.3. CPT Interpretation Method 59
0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90
0.00 0.05 0.10 0.15 0.20 0.25
Penetration Force in KN
u2 in m
p’0=100 kPa OCR=1.0 OCR=2.0 OCR=3.0
Figure 4.4: FE CPT simulation results: penetration force vs. penetration depth u2 (material #158).
has a smaller diameter than the cone. The gap between the soil and the rod cannot be modelled easily, since the viscohypoplastic constitutive model does not work for zero effective pressure. The other difficulty which arises, when the interface behaviour of the cone should consider friction, is the resulting discontinuity along the tube (tube under-neath the cone no friction, cone with friction, rod no friction). This results in numerical difficulties and premature aborts of the calculations. Another difficulty when allowing for contact friction in large displacement analyses is the distortion of the soil elements at the contact. This holds especially true when large gradients of stress and displacement occur at the interface. These gradients demand a fine FE mesh at the contact, but the smaller the elements, the larger is their distortion. Figure 4.5 shows a blow-up of the deformed mesh at the interface.
In order to allow for friction as far as possible, the calculation results are corrected after-wards. The calculations yield total stresses and pore water pressures at the cone, when the soil is in a limit state. With this the effective pressure normal to the surface of the cone can be calculated, and together with an assumed angle of contact friction δ ≈ ϕc/3 an additional friction component can be added to the penetration force of the probe. The corrected cone penetration resistance is calculated with
qc, FE = P F/Ac· fc (4.1)
where P F is the penetration force from the FE calculation, Ac is the projected area of
Figure 4.5: Deformed FE mesh near the ”cone” after penetration.
the cone and fc= 1 + tan 60◦tan δ is a correction factor allowing for friction at the cone (with an aperture of 60◦) with the angle of contact friction δ.
Validation of the FE Model
The calibration chamber tests (cf. Sec. 4.2.3) were simulated with the aid of the FE model described above. The penetration rates in the simulations were the same as measured in the tests in order to allow for the influence of viscosity.
The comparison of the measurements qt, m with the numerical results qc, FE is presented in Table 4.2. In the last column the deviation of the calculation result is given in percent with respect to the measured value. Neglecting the one outlier (test #BC1 with ∆ = −44.3%)
∆ ∈ [−16.3; 5.7]% holds. Considering the accuracy of the actual tests and the simplicity of the FE model, this is a very satisfying result and the FE model is considered to be validated.
Simulation Results and Interpretation Method
With the aid of the verified FE model CPT simulations were performed with the eight sets of viscohypoplastic parameters of the cohesive materials examined in this thesis (cf.
Chapter 2.5, Tab. 2.4 and A.3). For each material nine simulations with OCR = 1.0, 2.0 and 3.0 and p0 = p00 = (10, 100 and 250) kPa were performed.
Figure 4.6a depicts the calculation results for the Lake Constance clay (material #162) in terms of (qt− p0) versus p00. As can be seen, there is a linear relationship between these
4.3. CPT Interpretation Method 61
Brick Clay (#578)
Test p0 e OCR ∆qc, FE ∆qt, m ∆
in kPa in kPa in kPa in %
#BC1 42 0.793 2.17 342 237 -44.3
#BC2 134 0.738 1.02 558 517 -7.9
#BC3 51 0.748 2.48 464 399 -16.3
Lake Constance Clay (#162)
Test p0 e OCR ∆qc, FE ∆qc, m ∆
in kPa in MPa in MPa in %
#LC3 ≈ 150 ≈ 0.48 2.7 3.3 3.5 5.7
#LC4 ≈ 250 ≈ 0.44 3.7 6.4 5.8 -10.3
Table 4.2: Comparison between measured and calculated CPT results.
quantities for each OCR, this can be expressed as
0
In Figure 4.6b the inclinations tan α of the three straight lines in (a) are plotted versus
OCR. As can be seen in the diagram.
tan α = a · OCRb (4.3)
is an appropriate fitting function. Inserting (4.3) in (4.2) yields the final equation for the interpretation of CPTU results, viz
(qt− p0) = a · p00· OCRb. (4.4) The model parameters a and b were determined for all investigated materials (Tab. C.1, App. C.2.1).
Since the determination of a and b requires FE simulations with the non-conventional FE code Abaqus and a subroutine containing the viscohypoplastic constitutive equation, which is both in general not available to the practitioner, correlations between the two model parameters and soil mechanical properties of the material were searched for. It turned out that there is a linear relationship between the model parameters and Butter-field’s [6] compression and swelling index (λ and κ), the critical friction angle ϕc (Fig.
4.7), the consistency limits wL and wP, the plasticity index IP and the clay content CC (Fig. 4.8). The corresponding fitting functions are given in (4.5) and (4.6) together with the respective standard derivation s.
a = 6.3944 − 32.7586 · λ (s = 0.236) (4.5)
= 5.3531 − 105.981 · κ (s = 0.400)
= 11.0691 · ϕc− 1.67893 (s = 0.399)
= 6.4918 − 6.46535 · wL (s = 0.387)
= 8.3600 − 22.0996 · wP (s = 0.526)
= 5.6129 − 8.68667 · IP (s = 0.397)
= 5.6411 − 4.12652 · CC (s = 0.313)
b = 0.9425 − 1.27396 · λ (s = 0.021) (4.6)
= 0.9118 − 4.70601 · κ (s = 0.008)
= 0.4242 · ϕc+ 0.631469 (s = 0.025)
= 0.9532 − 0.266904 · wL (s = 0.019)
= 1.0186 − 0.85804 · wP (s = 0.028)
4.3. CPT Interpretation Method 63
0.04 0.06 0.08 0.10 0.12 0.14
a, b
0.00 0.01 0.02 0.03
a, b
0.30 0.40 0.50 0.60 0.70
a, b
ϕc in rad a
b
(c)
Figure 4.7: Correlation between the model parameters and (a) λ, (b) κ and (c) ϕc.
The actual interpretation works as follows: For known values of qt(z) from CPTU, p00(z) ≈ (γ0z·(1+2K0))/3 (where γ0is the specific buoyancy weight of the soil and K0the coefficient of earth pressure at rest), initial pore water pressure u0(z) from field measurements, p0(z) = p00(z) + u(z), the viscohypoplastic parameters λ and e100 (cf. Chap. 2.3) and the two model parameters a and b from numerical simulations or the correlations given above the void ratio e can be calculated from
0.50
0.20 0.30 0.40 0.50 0.60 0.70 0.80
a, b
0.00 0.10 0.20 0.30 0.40 0.50
a, b
0.00 0.20 0.40 0.60 0.80 1.00
a, b (d) clay content CC.
OCR =
Influence of the degree of saturation
4.3. CPT Interpretation Method 65
All simulations were carried out with a degree of saturation Sr = 0.99. Sr controls the bulk modulus ke of the pore fluid (water and gas) via
ke= kw· kg
Sr· kg + (1 − Sr) · kw (4.9) where kw ≈ 2.2 GPa is the bulk modulus of water, kg = pw+ pa of the pore gas, pw is the pore water pressure and pa≈ 100 kPa the atmospheric pressure.
Comparative calculations with different degrees of saturation and initial states were per-formed for one material (Fig. 4.9). With decreasing Sr and therewith ke the total pene-tration resistance qt− p0 increases, because a lower bulk modulus of the pore fluid leads to lower excess pore water pressures during shearing. The difference between calculations with Sr = 0.99 and 1.0 averages 1.7% (same holds for difference between Sr = 0.97 and 0.99). Since only a minor influence of Sr on qt− p0 was found, Sris not further considered in this study.
0 500 1000 1500 2000 2500
0 50 100 150 200 250 qt-p0 in kPa
p’0 in kPa Sr
0.97 0.98 0.99 1.00
OCR = 1.0 OCR = 1.5 OCR = 2.0
OCR = 2.5
Figure 4.9: Comparative simulations with different degrees of saturation Sr (material
#578).