Existing methods for determining c r from consolidation test with a central

Một phần của tài liệu A comparative study on the horizontal coefficient of consolidation (Cr) obtained from lab and field tests (Trang 28 - 38)

2.3.1 Root t method [11]

Introduction 2.3.1.1

The method proposed based on the equation for the equal vertical strain condition [4]. In Eq. (2.3) have Ur = f[Tr, F(n)] then the author can find Tr = f[Ur, F(n)].

(n) ln(1 U ) 8

r r

T  F

(2.21) T90 is the time factor at 90% average degree of consolidation so the value of T90 can calculated by F(n), Ur.

Berry (1969) commented that all the curve show linear portions between about 20%

- 60% average degree of consolidation. Thus a straight line is drawn through the experimental volume change –t0.5 results between about 20% to 60% consolidation, and a second line is then constructed having an abscissa 1.17 time that of the first [11].

The radial (horizontal) coefficient of consolidation is determined in this case:

2 90

90 e r

c T d

t

(2.22) The procedure for determine cr

2.3.1.2

The steps for determining the radial (horizontal) coefficient are the same as described in section.2.2.1.2

Evaluation of the method 2.3.1.3

Advantages

- This method is easy to practice for all engineers.

16

- Identify a straight line within Ur = 20% to Ur = 60% on the curve.

- This method does not need to find value 0 and 100. Disadvantages

- Straight line depends on the subjective evaluation of data processing engineers

2.3.2 Matching log (de2/t) and Ur method [12]

Introduction 2.3.2.1

The method proposed based on the equation for the equal vertical strain condition [4]. This approach solves the equation of Eq. (2.3) to find the dependence of Tr on Ur & F(n) then replaces Tr = f[Ur, F(n)] into Eq.(2.4).

Thus, the radial (horizontal) coefficient of consolidation (cr) is determined

2 8

( ) ln(1 U )

e r

r

d c

t  F n

 (2.23) The procedure for determine cr

2.3.2.2

- Step 1: Plot log (de2/t) versus Ur curve considering the δ - t data on Figure 2.7.

- Step 2: Identify a zone where the experimental curve is parallel to the theoretical curves

- Step 3: Using graphical or Eq. (2.23) can be determined cr.

Figure 2.7. Theoretical log(de2/t) versus Ur curves [12]

17 Evaluation of the method

2.3.2.3 Advantages

- cr can determine easily by the graph.

Disadvantages

- Matching between theoretical and experimental curve does not always occur - The variable Ur needs to determine exact values for 0 &100.

2.3.3 Inflection point method [13]

Introduction 2.3.3.1

The method was developed based on [13] and [4]. Eq. (2.3) can show the relationship Ur = f[log (Tr)].

According to the mathematical definition, the value of Ur maximum when the derivative d(Ur)/dlog Tr the maximum.

Figure 2.8. (a) Theretical Ur - log Tr curve and (b) d(Ur)/dlog Tr plot [13]

In this case, The degree of consolidation at the inflection point also the same for all the curves at Ur = Ur,inf = 63,21% with maximum derivative .

Thus, the value of Ur = Ur,inf = 63,21% can be calculated by Eq. (2.3)

18

r,inf

(n)ln(1 0.631) (n)

8 8

F F

T    

(2.24) The radial (horizontal) coefficient of consolidation is determined in this case:

2

inf

( ) 8

e r

d c F n

t

(2.25) The procedure for determine cr

2.3.3.2

- Step 1: Plot (Ur - log t) to t then finding time at max value (Ur - log t). This is the value tinf.

- Step 2: cr can be determined cr by Eq.(2.25) Evaluation of the method

2.3.3.3 Advantages

- 0, 100 does not need to be identified.

- In this method, the author finds tinf value.

Disadvantages

- There is no method yet to find tinf from Experimental data.

- The accuracy of results depends on the time distance between measurement results.

2.3.4 Non-graphical method [14]

Introduction 2.3.4.1

The method proposed based on the equation for the equal vertical strain condition [4].

The laboratory time (t) – compression r is described by the equation.

100 0

r Ur

   

(2.26) Ur is change in Eq. (2.26) from [4].

 100 ,0 2 0

1 exp 8 ( )

r

r r

e

C t F n d

         

 

  (2.27)

Combine Eq of (2.27) and constant values of de and 100, 0. F(n) can be found by matching between theoretical and experimental curve. Solve r – t curve can find cr.

19 The procedure for determine cr 2.3.4.2

Finding cr with100, 0, F(n), de and r – t curve by A source code or program on Eq.

(2.27).

Evaluation of the method 2.3.4.3

Advantages

- Use a coding program to resolve results independent of the implementer.

Therefore, the value has high accuracy.

- Results processing time is fast.

Disadvantages

- Matching between theoretical and experimental curve does not always occur - The value depends on the data range.

- Eq of (2.27) variable Ur needs to determine the exact values for 0 and 100. 2.3.5 Log - log method [15]

Introduction 2.3.5.1

The value of o can be calculated by selecting two time – settlement in the range Ur < 20% at experimental data(1, t1) & (2, t2).

2 2

1 1

o o

t t

 

 

    

     

  (2.28) Two straight lines depicted in Figure 2.9. The radial (horizontal) coefficient of consolidation is determined in this case:

2 ,66

66

(T )r e

r

c d

t

(2.29) The procedure for determine cr

2.3.5.2

- Step 1: Calculate the initial compression (0) using Eq. (2.28) from the time – compression data by choosing two points in the early stages of consolidation. The value of 0 can be calculated by selecting two time – settlement data within Ur < 20%.

20

- Step 2: Plot the time t – corrected settlement ( – 0) in a log – log plot.

- Step 3: Identify the initial linear portion and draw line.

- Step 4: Identify the linear secondary compression portion by drawing a line and extending it to intersect the initial straight line. The time at the point of intersection (t66) corresponds to a degree of consolidation of 66%.

- Step 5: cr can be determined by Eq. (2.29)

Figure 2.9. Log( - 0) versus log t plot [15]

Evaluation of the method 2.3.5.3

Advantages

- This method can determine 0 & 66.

- Methods Inheriting advantages of graphical method.

Disadvantages

- From Experimental data the value of 0 within Ur < 20% is not constant 2.3.6 Steepest tangent fitting method [16]

Introduction 2.3.6.1

The method Inflection point in Section 2.3.3 has disadvantages, Inflection point is difficult to determine exactly with experimental data. Vinod (2010) found a straight line through an Inflection point [16].

The equation of tangent through Inflection point on the semi-log graph (Figure 2.10) is determined:

 = b - alog(t)

(2.30)

21

where a, b = constant and (t, ) value of experimental data.

One log cycle, the author chooses value (1, t1), (2, t2) on the condition (t1 = 10 time), (t2 = 100 time) & (1 - 2 = h). Substituting (1, t1), (2, t2) into Eq. (2.30)

1 = ba & 2 = b – 2a Thus 1 - 2 = a = h

(2.31) Put a from Eq. (2.31).

 = b - hlog(t)

(2.32) P(o, to) is the corrected initial experimental data (Figure 2.10). Put b with P(o, to) from Eq. (2.32).

 = [0 + hlog(t0)]- alog(t) = 0 + hlog(t0 / t)

(2.33)

Figure 2.10. Steepest tangent fitting method for determination of cr

Similarly, a straight line through Inflection point on Ur-log Tr and d(Ur) /dlog Tr as shown Figure 2.8. Function for tangent on Ur - log Tr.

Ur = c - Slog(Tr)

(2.34) where c is constant and (Tr, Ur) value of predicted curve.

The value of S is defined by Section 2.3.3.

r,inf

0.847 63.2%

(log )

r r

S dU

d T U

 

 (2.35) Similarly, The corrected initial P(0, t0, Tr,0, Ur,0) is given by Ur,0 = c - Slog(T0) Substituting b with P(0, t0, Tr,0, Ur,0)

Ur = [Ur,0 + Slog(Tr,0)] - Slog(Tr) = Ur,0 + Slog(Tr,0 / Tr)

(2.36) Then dial reading  corresponding to x % consolidation

22 (Ur - Ur,0) / S = x / S = log(Tr,0 / Tr) = log(t0 / t)

(2.37) Eq. (2.37) and put log(to / t) from Eq. (2.33). Thus, function for steepest tangent is described as follows:

x = 0 + h x / S

(2.38) Author (Vinod, 2010) has determined the value of o as follows:

1 1

0 1 2

2 2

t / 1 t

t t

        

    (2.39) The procedure for determine cr

2.3.6.2

- Step 1: Plot the dial reading against time on semi log graph as show in Figure 2.10.

- Step 2: Determine o in Eq. (2.39)

- Step 3: Draw a tangent PQ to the steepest part of the consolidation curve.

- Step 4: Find h, which is the slope of the tangent PQ.

- Step 5: Find x use Eq.(2.38)

- Step 6: cr is calculated using Eq. (2.35).

Evaluation of the method 2.3.6.3

Advantages

- This method finds the 0 values.

- Only conduct experiments to Ur = 60%.

- Overcoming method disadvantages Inflection point.

Disadvantages

- Experimental data the value of 0 is not constant.

2.3.7 Log t method [17]

Introduction 2.3.7.1

Put Ur = f(0, , 100) from Eq. (2.3)

23

 ,t

0

100 0

1 exp 8Tr F n

 

 

 

     

    

    (2.40)

Replace t time in t1, t2 and 2t1 = t2

 ,1  

1 0

1

100 0

1 exp 8Tr 1 exp F n A

 

 

 

      

    

    (2.41)

 ,2  

2 0

2

100 0

1 exp 8Tr 1 exp F n A

 

 

 

      

    

   

Substituting A1, A2 with Eq.(2.42)

 1 1 0 100 1

100 0 100 0

exp A 1    

   

   

     (2.42)

 2 2 0 100 2

100 0 100 0

exp A 1    

   

   

    

Then

100 1

1

100 0

ln

A  

 

  

    (2.43)

100 2

2

100 0

ln

A  

 

  

   

Substituting A1,/A2 with Eq.(2.43)

100 1

100 0

2

1 100 2

100 0

ln

2 ln

A A

 

 

 

 

  

  

 

 

  

  

  (2.44)

In Eq. (2.44)

2

100 1 100 2 100 2

100 0 100 0 100 0

ln   2ln   ln  

     

        

 

        

      (2.45)

2

100 1 100 2

100 0 100 0

   

   

 

 

  

   

Then

 

2

100 1 2 1

100 2

(2 )

o

   

  

   

    (2.46)

24

Tr relationship fits approximately as a straight line within 0 < Ur < 20%.

The initial compression may be also determined graphically.

If the data points are selected such that t2 = 2t1, then Eq. (2.28) can show that

2 – 1 = 1 – 0

(2.47) The radial (horizontal) coefficient of consolidation is determined in this case:

2 50 50

(T )r e

r

c d

t

(2.48) The procedure for determine cr

2.3.7.2

- Step 1: Plot the time (t) – settlement () in a log t –  plot.

- Step 2: Draw a tangent through the inflection point.

- Step 3: identify the asymptotic secondary compression portion, draw a line, and extend it to intersect the tangent line. The point of intersection corresponds to a degree of consolidation of 100 % (100).

- Step 4: The value of 0 can be obtained using Eq. (2.28) or graphically using Eq.(2.46).

- Step 5: cr is calculated using Eq. (2.48).

Evaluation of the method 2.3.7.3

Advantages

- The value 0 can determine in this method.

- Only conduct experiments to Ur = 60%.

- Overcoming method disadvantages Inflection point.

Disadvantages

- Experimental data the value of 0 is not constant.

2.3.8 Full match method [10]

This method is identical for both CD & PD. The method is mentioned in the section (2.2.3)

The radial (horizontal) coefficient of consolidation is determined in this case:

25

2

8

n e r

c F d

Một phần của tài liệu A comparative study on the horizontal coefficient of consolidation (Cr) obtained from lab and field tests (Trang 28 - 38)

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