## LESSON 24. Determination of Coefficient of Consolidation

24.1 Taylor’s Square Root of Time Fitting Method

From the oedometer test (explained in Lesson 23) the dial reading (settlement) corresponding to a particular time is measured.  From the measured data, dial reading vs $\sqrt {Time}$ graph can be drawn (as shown in Figure 24.1). A straight line can be drawn passing through the points on initial straight portion of the curve (as shown in Figure 24.1). The intersection point between the straight line and the dial reading axis is denoted as R0 which is corrected zero reading i.e U = 0%.  Starting from R0, draw another straight line such that its abscissa is 1.15 times the abscissa of first straight line. The intersection point between the second straight line and experimental curve represents the R90 and corresponding $\sqrt {{t_{90}}}$ is determined. Thus, the time required (t90) for 90% consolidation is calculated. The Coefficient of consolidation (cv) is determined as:

${c_v}={{{T_v}{H^2}} \over t}$                    (24.1)

where H is the thickness of the soil sample, t is required time. Tv is the vertical time factor and can be determine as:

${T_v}=\left({{\pi\over 4}}\right){U^2}\quad \quad if\;U \le 60\%$   %                (24.2)

${T_v}=1.781 - 0.933{\log _{10}}(100 - U)\quad \quad if\;U > 60\%$   %               (24.3)

Fig. 24.1. Taylor’s Square Root of Time Fitting Method.

24.2 Casagrande’s Logarithm of Time Fitting Method

From the oedometer test (explained in Lesson 23) the dial reading (settlement) corresponding to a particular time is measured.  From the measured data, dial reading vs time graph can be drawn (as shown in Figure 24.2). Select two points t1 and t2 in initial part of the curve such that t2 = 4 t1. The points corresponding to the chosen times are marked on the curve. The vertical distance (z) between the two points on the curve is measured. Select another point R0 such that the vertical distance between that point and point on the curve corresponding to the t1 time is also z. Ris corrected zero reading i.e U = 0%. Determine the U=100% line by drawing two tangents form the straight portion of the curve as shown in Figure 24.2. Once U=0% and 100% lines are identified, U= 50% line is also determined by choosing the middle point between the U=0% and 100% lines. Time (t50) corresponding to the 50% degree of consolidation is determined from the curve. The Coefficient of consolidation (cv) is determined from Eq.(24.1).

Fig. 24.2. Casagrande’s Logarithm of Time Fitting Method.

Problem 1

Determine the total settlement of two layered soil system as shown in Figure 24.3. A strip loading of intensity 50 kN/m3 of width 3 m is applied at the ground surface. The water table is at a depth of 1.0 m below ground level.

Fig. 24.3. Settlement calculation of two layered soil system.

Solution:

The thickness of the Layer I and II is 3m and 2.5m, respectively. Choose one point at the middle of each normally consolidated clay layer. Here point A is chosen at a depth of 1.5 m from the ground level (if loading is applied at any depth from the ground level, then the point in the layer I will be chosen as middle of the soil layer in between the loading depth and top of the layer II. In layer II, point will be chosen at the middle of the layer II). The point B is chosen at a depth of 1.25m from the top of the layer II (or 4.25 m below the ground level).

Layer I

At point A,

σ'v0 = 18 × 1 + (1.5 – 1) × (20 – 10) = 23 kN/m2

Δσ'v = 50 × 3 /(3+1.5) = 33.33 kN/m2 (taking 1:2 distribution of loading)

H1 = 3m (if loading is applied at a depth, then H1 will be taken as total thickness of layer I minus the depth of the loading).

${S_1}={{{C_c}} \over {1 + {e_0}}}{H_1}{\log _{10}}\left( {{{{{\sigma '}_{v0}} + \Delta {{\sigma '}_v}} \over {{{\sigma '}_{v0}}}}} \right) = 0.12 \times 3 \times {\log _{10}}\left( {{{23 + 33.33} \over {23}}} \right)=140mm$

Layer II

At point B,

σ'v0 = 18 × 1 + 2 × (20 – 10) + 1.25 (19-10) = 49.25 kN/m2

Δσ'v = 50 × 3 /(3+4.25) = 20.69 kN/m2 (taking 1:2 distribution of loading)

H2 = 2.5m (if loading is applied at a depth, then H2 will be taken as total thickness of layer II).

${S_2}={{{C_c}} \over {1 + {e_0}}}{H_2}{\log _{10}}\left( {{{{{\sigma '}_{v0}} + \Delta {{\sigma '}_v}} \over {{{\sigma '}_{v0}}}}} \right) = 0.16x2.5x{\log _{10}}\left( {{{49.25 + 20.69} \over {49.25}}} \right)=60.93mm$

Total settlement = S1+S2= 140 +60.93 = 200.93 mm

References

Ranjan, G. and Rao, A.S.R. (2000). Basic and Applied Soil Mechanics. New Age International Publisher, New Delhi, India.