The Engineering of Foundations Lecture 3 SPT borehole log Cone penetration test Indirect interpretation of SPT results Sand Clay 10/5/2013 Al-Qadissya University

A sample of SPT log is shown in the Fig. below. SPT borehole log A sample of SPT log is shown in the Fig. below. Fig 3.1 Borehole log of SPT test 10/5/2013

Borehole log of SPT (continued) SPT borehole log Borehole log of SPT (continued) 10/5/2013

Cone Penetration Test (CPT) As in the Fig., the cone penetrometer is a cone with a base area of 10 cm2 and cone angle 600 that is attached to the rod. An outer sleeve encloses the rod. Fig. 3.2 Electric cone The cone is driven at a rate of 2 cm/sec and the cone resistance and sleeve friction are measured separately. 10/5/2013

CPT Advantages of the cone Very useful for soil profiling Data generation during test Cheap in comparison with boring Provides data can be used to estimate shear strength, bearing capacity and consolidation characteristics of soil Other attachments can be incorporated for further data acquisition Electric cone 10/5/2013

A typical CPT log is shown in the Figure. Fig 3.3 CPT log for clay 10/5/2013

Interpretation of SPT results There are two approaches to the SPT results for foundation design: Direct interpretation in which SPT results are directly related to foundation capacity. Indirect interpretation in which SPT results are related to soil properties and then with foundation capacity. Direct interpretation will be discussed later. 10/5/2013

Indirect interpretation of SPT results Sand It is assumed that Nspt increases with increase of relative density, Dr, and confining stress. The following correlation rely on blow count N, normalized for the vertical effective stress. It applies strictly to NC sand for which K0 range 0.38-0.50. (3.1) pA = 100 kPa 10/5/2013

Indirect interpretation of SPT results The Engineering of Foundations Indirect interpretation of SPT results A general equation is proposed by Salgado et al. (1997) is as follows: (3.2) Skempton (1986) proposed the following expression for relative density in terms of N60 35%≤ Dr ≤ 85% 50kPa ≤ σv ≤ 250 kPa (3.3) 27 ≤ A ≤ 46 B≈ 27 C = (1+2k0)/(1+2k0.NC) (3.4) 10/5/2013 Al-Qadissya University

Indirect interpretation of SPT results indirect correlation between Nspt and the friction angle is also proposed as shown in the Fig. below Fig 3.4 correlation SPT data and friction angle 10/5/2013

Indirect interpretation of SPT results Example: refer to the homework 1, the sandy deposit is located near a stream and is known to be NC. The average unit weight of the sand both below and above the water table is 20 kN/m3. Obtain the stress normalized blow count value at the depth of 2 m and 10 m. Solution: the energy corrected at 2 m depth equal to 2.7 and 18 at depth of 10 m. Assume WT at level of 2 m. σv = 2*20 = 40 kPa, σv= 10*20-8*9.8 = 122 kPa The stress normalized blow counts is N1(2m) = 2.7(100/40)^0.5 = 4.3 10/5/2013

Indirect interpretation of SPT results N1 (10m) = 18*(100/122)^0.5 = 16 Clay The SPT based correlations for clay should be treated with caution due to many uncertainties in the SPT shear strength of clay depends on blow counts which depend on loading rate Blow count depends on clay sensitivity which is the ratio of strength of undisturbed to strength of disturbed. In organic or very soft soils the SPT split spoon may go down without any hammer blow There are other factors affect the accuracy of SPT 10/5/2013

Indirect interpretation of SPT results Clayton (1990) proposed the following correlation between SPT results and undrained shear strength. (3.5) (3.6) PA = 100 kPa, f1, f2 are given in the Figure provided mv = compressibility 10/5/2013

Indirect interpretation of SPT results Example: for clay at 10 m depth, an N60 count was determined to be equal to 30. this clay has LL = 53% and PL = 20%. The unit weight of the clay is 16 kN/m3. Determine the undrained shear strength at 10 m. what is the value of su/σv at 10 m if water table is at surface. Evaluate OCR and comment on the result. Solution: Apply Eq. 3.5, PI = 33%, f1 = 0.045 (Fig ) Su = 0.82*0.045*30*100 = 111 kPa Vertical effective stress σv = (16-9.8)*10 = 62 kPa Su/ σv = 111/62 = 1.8 10/5/2013

Indirect interpretation of SPT results Using the correction exists for Su for NC clays in terms of the PI (Su/ σv )NC = 0.11*0.0037*PI = 0.11*0.0037*33 = 0.23 The value of Su/ σv we calculated at 10 m depth is higher than 0.23 suggesting that the clay is over-consolidated we can estimate OCR from: (Su/ σv)OC /(Su/ σv )NC = OCR ^0.8 1.8/0.23 = 7.8 OCR = 13 10/5/2013