What Is the Dial Reading in Taylor's Square Root of Time Method
24.ane Taylor'due south Square Root of Time Fitting Method
From the oedometer test (explained in Lesson 23) the dial reading (settlement) respective to a particular fourth dimension is measured. From the measured data, dial reading vs \[\sqrt {Time}\] graph can exist drawn (equally shown in Effigy 24.one). A straight line can be drawn passing through the points on initial straight portion of the bend (equally shown in Effigy 24.i). The intersection bespeak between the straight line and the dial reading axis is denoted as R0 which is corrected zilch reading i.eastward U = 0%. Starting from R 0, describe another straight line such that its abscissa is 1.15 times the abscissa of get-go direct line. The intersection signal betwixt the 2d directly line and experimental curve represents the R 90 and respective \[\sqrt {{t_{90}}}\] is adamant. Thus, the time required (t 90) for 90% consolidation is calculated. The Coefficient of consolidation (c v) is determined every bit:
\[{c_v}={{{T_v}{H^two}} \over t}\] (24.1)
where H is the thickness of the soil sample, t is required time. T v is the vertical fourth dimension gene and can be make up one's mind as:
\[{T_v}=\left({{\pi\over iv}}\correct){U^ii}\quad \quad if\;U \le 60\% \] % (24.2)
\[{T_v}=1.781 - 0.933{\log _{ten}}(100 - U)\quad \quad if\;U > 60\% \] % (24.3)
Fig. 24.1. Taylor'due south Square Root of Time Plumbing equipment Method.
24.ii Casagrande'due south Logarithm of Fourth dimension Fitting Method
From the oedometer test (explained in Lesson 23) the dial reading (settlement) respective to a particular fourth dimension is measured. From the measured data, dial reading vs time graph can be drawn (as shown in Effigy 24.2). Select two points t one and t ii in initial part of the curve such that t 2 = 4 t 1. The points respective to the chosen times are marked on the bend. The vertical distance (z) betwixt the two points on the bend is measured. Select another point R 0 such that the vertical altitude betwixt that point and betoken on the curve corresponding to the t one time is also z. R 0 is corrected null reading i.e U = 0%. Determine the U=100% line by cartoon two tangents class the straight portion of the curve equally shown in Figure 24.2. One time U=0% and 100% lines are identified, U= 50% line is also adamant by choosing the middle betoken between the U=0% and 100% lines. Fourth dimension (t 50) corresponding to the 50% caste of consolidation is determined from the curve. The Coefficient of consolidation (c5) is determined from Eq.(24.one).
Fig. 24.2. Casagrande's Logarithm of Time Fitting Method.
Problem 1
Determine the total settlement of two layered soil organisation as shown in Figure 24.3. A strip loading of intensity fifty kN/yardiii of width iii m is applied at the footing surface. The water tabular array is at a depth of 1.0 g below ground level.
Fig. 24.3. Settlement calculation of two layered soil system.
Solution:
The thickness of the Layer I and Two is 3m and 2.5m, respectively. Choose i point at the center of each ordinarily consolidated clay layer. Here betoken A is chosen at a depth of 1.v chiliad from the ground level (if loading is applied at any depth from the footing level, then the point in the layer I will be chosen as eye of the soil layer in betwixt the loading depth and top of the layer Ii. In layer Two, point will be chosen at the heart of the layer II). The signal B is called at a depth of 1.25m from the top of the layer II (or 4.25 k beneath the basis level).
Layer I
At point A,
σ'v0 = 18 × i + (ane.v – ane) × (20 – ten) = 23 kN/k2
Δσ'5 = 50 × 3 /(3+one.v) = 33.33 kN/g2 (taking 1:2 distribution of loading)
H ane = 3m (if loading is applied at a depth, and then H 1 will be taken every bit full thickness of layer I minus the depth of the loading).
\[{S_1}={{{C_c}} \over {1 + {e_0}}}{H_1}{\log _{10}}\left( {{{{{\sigma '}_{v0}} + \Delta {{\sigma '}_v}} \over {{{\sigma '}_{v0}}}}} \right) = 0.12 \times three \times {\log _{10}}\left( {{{23 + 33.33} \over {23}}} \correct)=140mm\]
Layer Ii
At betoken B,
σ'v0 = 18 × 1 + two × (20 – 10) + one.25 (xix-ten) = 49.25 kN/yard2
Δσ'v = 50 × 3 /(3+4.25) = xx.69 kN/m2 (taking 1:2 distribution of loading)
H 2 = 2.5m (if loading is applied at a depth, and then H 2 will be taken as total thickness of layer Ii).
\[{S_2}={{{C_c}} \over {1 + {e_0}}}{H_2}{\log _{ten}}\left( {{{{{\sigma '}_{v0}} + \Delta {{\sigma '}_v}} \over {{{\sigma '}_{v0}}}}} \correct) = 0.16x2.5x{\log _{ten}}\left( {{{49.25 + 20.69} \over {49.25}}} \right)=60.93mm\]
Total settlement = Due southane+Due south2= 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.
Suggested Readings
Ranjan, Yard. and Rao, A.S.R. (2000) Basic and Applied Soil Mechanics. New Age International Publisher, New Delhi, India.
Arora, Chiliad.R. (2003) Soil Mechanics and Foundation Technology. Standard Publishers Distributors, New Delhi, Republic of india.
Murthy V.Northward.S (1996) A Text Book of Soil Mechanics and Foundation Engineering science, UBS Publishers' Distributors Ltd. New Delhi, India.
Source: http://ecoursesonline.iasri.res.in/mod/page/view.php?id=125167
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