NPTEL - ADVANCED FOUNDATION ENGINEERING-1. Module 1. (Lecture 3).
GEOTECHNICAL PROPERTIES OF SOIL AND OF. REINFORCED SOIL. Topics.
NPTEL - ADVANCED FOUNDATION ENGINEERING-1
Module 1 (Lecture 3) GEOTECHNICAL PROPERTIES OF SOIL AND OF REINFORCED SOIL Topics 1.1 CAPILLARY RISE IN SOIL 1.2 CONSOLIDATIONS-GENERAL 1.3 CONSOLIDATION SETTLEMENT CALCULATION 1.4 TIME RATE OF CONSOLIDATION
CAPILLARY RISE IN SOIL When a capillary tube is placed in water, the water level in the tube rises (figure 1.15a). The rise is caused by the surface tension effect. According to figure 1.15a, the pressure at any point A in the capillary tube (with respect to the atmospheric pressure) can be expressed as 𝑢𝑢 = −𝛾𝛾𝑤𝑤 𝑧𝑧 ′ (for 𝑧𝑧 ′ = 0 to ℎ𝑐𝑐 ) And
𝑢𝑢 = 0
(for z ′ ≥ hc )
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Figure 1.15 Capillary rise In a given soil mass, the interconnected void spaces can behave like a number of capillary tubes with varying diameters. The surface tension force may cause in the soil to rise above the water table, as shown in figure 1.15b. The height the capillary rise will depend on the diameter of the capillary tubes. The capillary rise will decrease with the increase of the tube diameter. Because the capillary tube in soil has variable diameters, the height of capillary rise will be nonuniformly. The pore water pressure at any point in the zone of capillary rise in soil cause approximated as 𝑢𝑢 = −𝑆𝑆𝛾𝛾𝑤𝑤 𝑧𝑧′
[1.52]
Where
𝑆𝑆 = degree fo saturation of soil [equation (7)]
𝑧𝑧 ′ = distance measured above the water table CONSOLIDATION-GENERAL
In the field, when the stress on a saturated clay layer is increased-for exam by the construction of a foundation-the pore water pressure in the clay increase. Because the hydraulic conductivity of clays is very small, sometime be required for the excess pore water pressure to dissipate and the stress increase to be transferred to the soil skeleton gradually. According to figure 1.16 if ∆ a surcharge at the ground surface over a very large area, the increase of total structure ∆𝜎𝜎, at any depth of the clay layer will be equal to ∆𝑝𝑝, or ∆𝜎𝜎 = ∆𝑝𝑝
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Figure 1.16 Principles of consolidation However, at time 𝑡𝑡 = 0 (that is, immediately after the stress application), the excess pore water pressure at any depth, ∆𝑢𝑢, will equal ∆𝑝𝑝, or ∆𝑢𝑢 = ∆ℎ1 𝛾𝛾𝑤𝑤 = Δ𝑝𝑝
(at time 𝑡𝑡 = 0)
Hence the increase of effective stress at time 𝑡𝑡 = 0 will be Δ𝜎𝜎 ′ = Δ𝜎𝜎 − Δ𝑢𝑢 = 0
Theoretically, at time 𝑡𝑡 = ∞, when all the excess pore water pressure in the clay layer has dissipated as a result of drainage into the sand layers, Δ𝑢𝑢 = 0
at time 𝑡𝑡 = ∞)
Then the increase of effective stress in the clay layer is Δ𝜎𝜎 ′ = Δ𝜎𝜎 − Δ𝑢𝑢 = Δ𝑝𝑝 − 0Δ𝑝𝑝
This gradual increase in the effective stress in the claylayer will cause settlement over a period of time and is referred to as consolidation. Laboratory tests on undisturbed saturated clay specimens can be conducted (ASTM Test Designation D-2435) to determine the consolidation settlement caused by various incremental loadings. The test specimens are usually 2.5 in. (63.5 mm) in diameter and 1 in. (25.4 mm) in height. Specimens are placed inside a ring, with one porous stone at the top and one at the bottom of the specimen (figure 1.17a). Load on the specimen is then applied so that the total vertical stress is equal to 𝑝𝑝. Settlement readings for the specimen are taken for 24 hours. After that, the load on the specimen is doubled and settlement readings are taken. At all times during the test the specimen is kept under water. This procedure is continued until the desired limit of stress on the clay specimen is reached.
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Figure 1.17 (a) Schematic diagram of consolidation test arrangement; (b) 𝑒𝑒 − log 𝑝𝑝 curve for a soft clay from East St. Louis, Illinois
Figure 1.17 continued Based on the laboratory tests, a graph can be plotted showing the variation of the void ratio 𝑒𝑒 at the end of consolidation against the corresponding vertical stress 𝑝𝑝
NPTEL - ADVANCED FOUNDATION ENGINEERING-1 (semilogarithmic graph: 𝑒𝑒 on the arithmetic scale and 𝑝𝑝 on the log scale). The nature of variation of 𝑒𝑒 against log 𝑝𝑝 for a clay specimen is shown in figure 1.17b. After the desired consolidation pressure has been reached, the specimen can be gradually unloaded, which will result in the swelling of the specimen. Figure 1.17b also shows the variation of the void ratio during the unloading period. From the 𝑒𝑒 − log − 𝑝𝑝 curve shown in figure 1.17b, three parameters necessary for calculating settlement in the field can be determined.
1. The preconsolidation pressure, 𝑝𝑝𝑐𝑐 , is the maximum past effective overburden pressure to which the soil specimen has been subjected. It can be determined by using a simple graphical procedure as proposed by Casegrande (1936). This procedure for determining the preconsolidation pressure, with reference to figure 1.17b, involves five steps: a. Determine the point O on the 𝑒𝑒 − log 𝑝𝑝 curve that has the sharpest curvature (that is, the smallest radius of curvature). b. Draw a horizontal line OA. c. Draw a line OB that is tangent to the 𝑒𝑒 − log 𝑝𝑝 curve at O. d. Draw a line OC that bisects the angle AOB, e. Produce the straight-line portion of the 𝑒𝑒 − log 𝑝𝑝 curve backward to intersect OC. This is point D. the pressure that corresponds to point 𝑝𝑝 is the preconsolidation pressure, 𝑝𝑝𝑐𝑐 .
Natural soil deposits can be normally consolidated or overconsolidated (or preconsolidated). If the present effective overburden pressure 𝑝𝑝 = 𝑝𝑝𝑎𝑎 is equal to the preconsolidated pressure 𝑝𝑝𝑐𝑐 the soil is normally consolidated. However, if 𝑝𝑝𝑜𝑜 < 𝑝𝑝𝑐𝑐 , the sol is overconsolidated. Preconsolidation pressure (𝑝𝑝𝑐𝑐 ) has been correlated with the index parameters by several investigators. Stas and Kulhawy (1984) suggested that 𝑝𝑝 𝑐𝑐
𝜎𝜎𝑎𝑎
= 10(1.11−1.62𝐿𝐿𝐿𝐿)
[1.53a]
Where
𝜎𝜎𝑎𝑎 = atmospheric stress in derived unit 𝐿𝐿𝐿𝐿 = liquidity index
The liquidity index of a soil is defined as 𝐿𝐿𝐿𝐿 =
𝑤𝑤 −𝑃𝑃𝑃𝑃 𝐿𝐿𝐿𝐿_𝑃𝑃𝑃𝑃
Where
𝑤𝑤 = 𝑖𝑖𝑖𝑖 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 moisture content
[1.53b]
NPTEL - ADVANCED FOUNDATION ENGINEERING-1 𝐿𝐿𝐿𝐿 = liquid limit
𝑃𝑃𝑃𝑃 = plastic limit
Nagaraj and Murthy (1985) provided an empirical relation to calculate 𝑝𝑝𝑐𝑐 , which is as follows:
log 𝑝𝑝𝑐𝑐 =
↑ kN/m2
kN/m2 ↓
𝑒𝑒 1.122−� 𝑜𝑜 �−0.0463 log 𝑝𝑝 𝑜𝑜 𝑒𝑒 𝐿𝐿
0.188
[1.54]
Where
𝑒𝑒𝑜𝑜 = 𝑖𝑖𝑖𝑖 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 void ratio
𝑝𝑝𝑜𝑜 = 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 effective overburden pressure 𝑒𝑒𝐿𝐿 = void ratio of the soil at liquid limit 𝑒𝑒𝐿𝐿 = �
𝐿𝐿𝐿𝐿(%) 100
� 𝐺𝐺𝑠𝑠
[1.55]
The U. S. Department of the Navy (1982) also provided generalized relationships between 𝑝𝑝𝑐𝑐 , 𝐿𝐿𝐿𝐿 and the sensitivity of clayey soils (𝑆𝑆𝑡𝑡 ). This relationship was also recommended by Kulhawy and Mayne (1990). The definition of sensitivity is given in section. Figure 1.18 shows the relationship.
Figure 1.18 Variation of 𝑝𝑝𝑐𝑐 with LI (after U. S. Department of the Navy, 1982)
NPTEL - ADVANCED FOUNDATION ENGINEERING-1 2. The compression index, 𝐶𝐶𝑐𝑐 , is the slope of straight-line portion (latter part of the loading curve), or 𝑒𝑒 −𝑒𝑒
2 𝐶𝐶𝑐𝑐 = log 𝑝𝑝 1 −log = 𝑝𝑝 2
1
𝑒𝑒1 −𝑒𝑒2
𝑝𝑝 log � 2 �
[1.56]
𝑝𝑝 1
where 𝑒𝑒1 and 𝑒𝑒2 are the void ratios at the end of consolidation under stresses 𝑝𝑝1 and 𝑝𝑝2 , respectively
The compression index, as determined from the laboratory 𝑒𝑒 − log 𝑝𝑝 curve, will be somewhat different from that encountered in the field. The primary reason is that the soil remolds to some degree during the field exploration. The nature of variation of the 𝑒𝑒 − log 𝑝𝑝 curve in the field for normally consolidated clay is shown in figure 1.19. It is generally referred to as the virgin compression curve. The virgin curve approximately intersects the laboratory curve at a void ratio of 0.42𝑒𝑒𝑜𝑜 (Terzaghi and Peck, 1967). Note that 𝑒𝑒𝑜𝑜 is the void ratio of the clay in the field. Knowing the values of 𝑒𝑒𝑜𝑜 and 𝑝𝑝𝑐𝑐 you can easily construct the virgin curve and calculate the compression index of the virgin curve by using equation (56).
Figure 1.19 Construction of virgin compression curve for normally consolidated clay
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The value of 𝐶𝐶𝑐𝑐 can vary widely depending on the soil. Skempton (1944) has given am empirical correlation for the compression index in which 𝐶𝐶𝑐𝑐 = 0.009(𝐿𝐿𝐿𝐿 − 10)
[1.57]
Where 𝐿𝐿𝐿𝐿 = liquid limit
Besides Skempton, other investigators have proposed correlations for the compression index. Some of these correlations are summarized in table 14. 3. The swelling index, 𝐶𝐶𝑠𝑠 , is the slope of the unloading portion of the 𝑒𝑒 − log 𝑝𝑝 curve. In figure 1.17b, it can be defined as 𝐶𝐶𝑠𝑠 =
𝑒𝑒3 −𝑒𝑒4 𝑝𝑝 log � 4 �
[1.58]
𝑝𝑝 3
In most cases the value of the swelling index (𝐶𝐶𝑠𝑠 ) is 14 to 15 of the compression index. Flowing are some representative values of 𝐶𝐶𝑠𝑠 /𝐶𝐶𝑐𝑐 for natural soil deposits. The swelling index is also referred to as the recompression index. 𝐶𝐶𝑠𝑠 /𝐶𝐶𝑐𝑐
Description of soil Boston Blue clay
0.24-0.33
Chicago clay
0.15-0.3
New Orleans clay
0.15-0.28
St. Lawrence clay
0.05-0.1
Table 14 Correlations for Compression Index Reference Azzouz, Krizek, and Corotis (1976)
Correlation 𝐶𝐶𝑐𝑐 = 0.01 𝑤𝑤𝑛𝑛
(Chicago clay)
𝐶𝐶𝑐𝑐 = 0.208 𝑒𝑒𝑜𝑜 + 0.0083 𝐶𝐶𝑐𝑐 = 0.0115 𝑤𝑤𝑛𝑛
(Chicago clay)
(organic soils, peat)
𝐶𝐶𝑐𝑐 = 0.0046(𝐿𝐿𝐿𝐿 − 9)
(Brazillian clay)
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Rendon-Herrero (1980)
Nagaraj and Murthy (1985) Wroth and Wood (1978) Leroueil, Tavenas, and LeBihan (1983)
𝐶𝐶𝑐𝑐 =
0.141𝐺𝐺𝑠𝑠12
1 + 𝑒𝑒𝑜𝑜 2.38 � � 𝐺𝐺𝑠𝑠
𝐿𝐿𝐿𝐿 𝐶𝐶𝑐𝑐 = 0.2343 � � 𝐺𝐺 100 𝑠𝑠 𝑃𝑃𝑃𝑃 𝐶𝐶𝑐𝑐 = 0.5𝐺𝐺𝑠𝑠 � � 100
Note: 𝐺𝐺𝑠𝑠 = specific gravity of soil solids 𝐿𝐿𝐿𝐿 = liquid limit
𝑃𝑃𝑃𝑃 = plasticity index 𝑆𝑆𝑡𝑡 = sensitivity
𝑤𝑤𝑛𝑛 = natural moisture content
The swelling index determination is important in the estimation of consolidation settlement of overconsolidated clays. In the field, depending on the pressure increase, an overconsolidated clay will follow an e-log 𝑝𝑝 path 𝑎𝑎𝑎𝑎𝑎𝑎, as shown in figure 1.20. Note that point 𝑎𝑎 with coordinates of 𝑝𝑝𝑜𝑜 and 𝑒𝑒𝑜𝑜 corresponds to the field condition before any pressure increase. Point 𝑏𝑏 corresponds to the preconsolidation pressure (𝑝𝑝𝑐𝑐 ) of the clay. Line 𝑎𝑎𝑎𝑎 is approximately parallel to the laboratory unloading cure 𝑐𝑐𝑐𝑐 (Schmertmann, 1953). Hence, if you know 𝑒𝑒𝑜𝑜 , 𝑝𝑝𝑜𝑜 , 𝑝𝑝𝑐𝑐 , 𝐶𝐶𝑐𝑐 , and 𝐶𝐶𝑠𝑠 , you can easily construct the field consolidation curve.
NPTEL - ADVANCED FOUNDATION ENGINEERING-1
Figure 1.20 Construction of field consolidation curve for over consolidated clay Nagaraj and Murthy (1985) expressed the swelling index as 𝐿𝐿𝐿𝐿
𝐶𝐶𝑠𝑠 = 0.0463 �100 � 𝐺𝐺𝑠𝑠
[1.59]
It is essential to point out that any of the empirical correlations for 𝐶𝐶𝑐𝑐 and 𝐶𝐶𝑠𝑠 given in the section are only approximate. It may be valid for a given soil for which the relationship was developed but may not hold good for other soils. As an example, figure 1.21 shows the plots of 𝐶𝐶𝑐𝑐 and 𝐶𝐶𝑠𝑠 with liquid limit for soils from Richmond, Virginia (Martin et al., 1985). For these soils,
NPTEL - ADVANCED FOUNDATION ENGINEERING-1
Figure 1.21 Variation of 𝐶𝐶𝑐𝑐 and 𝐶𝐶𝑠𝑠 with liquid limit for soils from Richmond, Virginia (after Martin et al., 1995) 𝐶𝐶𝑐𝑐 = 0.0326(𝐿𝐿𝐿𝐿 − 43.4)
[1.60]
And 𝐶𝐶𝑠𝑠 = 0.00045(𝐿𝐿𝐿𝐿 + 11.9)
[1.61]
The 𝐶𝐶𝑠𝑠 / 𝐶𝐶𝑐𝑐 ratio is about 251 ; whereas, the typical range is about15 to 101 . CONSOLIDATION SETTLEMENT CALCULATION The one-dimensional consolidation settlement (caused by an additional load) of a clay layer (figure 1.22a) having a thickness 𝐻𝐻𝑐𝑐 may be calculated as
NPTEL - ADVANCED FOUNDATION ENGINEERING-1 Δ𝑒𝑒
𝑆𝑆 = 1+𝑒𝑒 𝐻𝐻𝑐𝑐 𝑜𝑜
[1.62]
Figure 1.22 One-dimensional settlement calculation: (b) is for equation (64); (c) is for equations (66 and 68) Where 𝑆𝑆 = settlement
Δ𝑒𝑒 = total change of void ratio caused by the additional load application 𝑒𝑒𝑜𝑜 = the void ratio of the clay before the application of load Note that Δ𝑒𝑒
1+𝑒𝑒𝑜𝑜
= 𝜀𝜀𝑣𝑣 = vertical strain
For normally consolidated clay, the field 𝑒𝑒 − log 𝑝𝑝 curve will be like the one shown in figure 1.22b. If 𝑝𝑝𝑜𝑜 = initial average effective overburden pressure on the clay layer and Δ𝑝𝑝 = average pressure increase on the clay layer caused by the added load, the change of void ratio caused by the load increase is
NPTEL - ADVANCED FOUNDATION ENGINEERING-1 Δ𝑒𝑒 = 𝐶𝐶𝑐𝑐 log
𝑝𝑝 𝑜𝑜 +Δ𝑝𝑝
𝐶𝐶 𝐻𝐻
𝑝𝑝 𝑜𝑜 +Δ𝑝𝑝
[1.63]
𝑝𝑝 𝑜𝑜
Now, combining equations (62 and 63) yields 𝑐𝑐 𝑐𝑐 𝑆𝑆 = 1+𝑒𝑒 log 𝑜𝑜
[1.64]
𝑝𝑝 𝑜𝑜
For overconsolidated clay, the field 𝑒𝑒 − log 𝑝𝑝 curve will be like the one show figure 1.22c. In this case, depending on the value of Δ𝑝𝑝, two conditions may at. First, if 𝑝𝑝𝑜𝑜 + Δ𝑝𝑝 < 𝑝𝑝𝑐𝑐 , Δ𝑒𝑒 = 𝐶𝐶𝑠𝑠 log
𝑝𝑝 𝑜𝑜 +Δ𝑝𝑝
[1.65]
𝑝𝑝 𝑜𝑜
Combining equations (62 and 65) gives 𝐻𝐻 𝐶𝐶
𝑐𝑐 𝑠𝑠 𝑆𝑆 = 1+𝑒𝑒 log 𝑜𝑜
𝑝𝑝 𝑜𝑜 +Δ𝑝𝑝
[1.66]
𝑝𝑝 𝑜𝑜
Second, if 𝑝𝑝𝑜𝑜 < 𝑝𝑝𝑐𝑐 < 𝑝𝑝𝑜𝑜 + ∆𝑝𝑝, 𝑝𝑝
∆𝑒𝑒 = ∆𝑒𝑒1 + ∆𝑒𝑒2 = 𝐶𝐶𝑧𝑧 log 𝑝𝑝 𝑐𝑐 + 𝐶𝐶𝑐𝑐 log 𝑜𝑜
𝑝𝑝 𝑜𝑜 +Δ𝑝𝑝 𝑝𝑝 𝑜𝑜
[1.67]
Now, combining equations (62 and 67) yields 𝐶𝐶 𝐻𝐻
𝑝𝑝
𝐶𝐶 𝐻𝐻
𝑠𝑠 𝑐𝑐 𝑐𝑐 𝑐𝑐 𝑆𝑆 = 1+𝑒𝑒 log 𝑝𝑝 𝑐𝑐 + 1+𝑒𝑒 log 𝑜𝑜
𝑜𝑜
𝑜𝑜
𝑝𝑝 𝑜𝑜 +Δ𝑝𝑝 𝑝𝑝 𝑐𝑐
[1.68]
TIME RATE OF CONSOLIDATION In section we showed that consolidation is the result gradual dissipation of the excess pore water pressure from a clay layer. Pore water pressure dissipation, in turn, increases the effective stress, which induces settlement. Hence, to estimate the degree of consolidation of a clay layer at some time t after the load application, you need to know the rate of dissipation of the excess pore water pressure. Figure 1.23 shows a clay layer of thickness 𝐻𝐻𝑐𝑐 that has highly permeable sand layers at its top and bottom. Here, the excess pore pressure at any point at any time t after the load application is ∆𝑢𝑢 = (∆ℎ)𝛾𝛾𝑤𝑤 . For a vertical drainage condition (that is, in the direction of z only) from the clay layer, Terzaghi derived the following differential equation:
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Figure 1.23 (a) Derivation of equation (71); (b) nature of variation of ∆𝑢𝑢 with time 𝜕𝜕(∆𝑢𝑢) 𝜕𝜕𝜕𝜕
= 𝐶𝐶𝑣𝑣
𝜕𝜕 2 (∆𝑢𝑢)
[1.69]
𝜕𝜕𝜕𝜕 2
Where 𝐶𝐶𝑣𝑣 = coefficient of consolidation 𝐶𝐶𝑣𝑣 = 𝑚𝑚 Where
𝑘𝑘
𝑣𝑣 𝛾𝛾𝑤𝑤
=
𝑘𝑘
∆𝑒𝑒 𝛾𝛾 ∆𝑝𝑝 (1+𝑒𝑒 𝑎𝑎𝑎𝑎 ) 𝑤𝑤
[1.70]
𝑘𝑘 = hydraulic conductivity of the clay
∆𝑒𝑒 = total change of void ratio caused by a stress increase of ∆p 𝑒𝑒𝑎𝑎𝑎𝑎 = average void ratio during consolidation
𝑚𝑚𝑣𝑣 = volume coefficient of compressibility = ∆𝑒𝑒/[∆𝑝𝑝(1 + 𝑒𝑒𝑎𝑎𝑎𝑎 )]
Equation (69) can be solved to obtain ∆𝑢𝑢 as a function of time t with the following boundary conditions:
NPTEL - ADVANCED FOUNDATION ENGINEERING-1 1. Because highly permeable sand layers are located at 𝑧𝑧 = 0 and 𝑧𝑧 = 𝐻𝐻𝑐𝑐 , the excess pore water pressure developed in the clay at those points will be immediately dissipated. Hence ∆𝑢𝑢 = 0 at 𝑧𝑧 = 0 ∆𝑢𝑢 = 0 at 𝑧𝑧 = 𝐻𝐻𝑐𝑐 = 2𝐻𝐻 Where
𝐻𝐻 = Length of maximum drainage path (due to two-way drainage condition-that is, at the top and bottom of the clay) 2. At time 𝑡𝑡 = 0, ∆𝑢𝑢 = ∆𝑢𝑢𝑜𝑜 = initial excess pore water pressure after the load application With the preceding boundary conditions, equation (69) yields =∞ ∆𝑢𝑢 = ∑𝑚𝑚 � 0
2(∆𝑢𝑢 𝑜𝑜 ) 𝑀𝑀
𝑀𝑀𝑀𝑀
𝑠𝑠𝑠𝑠𝑠𝑠 � 𝐻𝐻 �� 𝑒𝑒 −𝑀𝑀
2 𝑇𝑇 𝑣𝑣
[1.71]
Where 𝑀𝑀 = [(2𝑚𝑚 + 1)𝜋𝜋]/2 𝑚𝑚 = an integer = 1, 2, … 𝑇𝑇𝑣𝑣 = nondimensional time factor = (𝐶𝐶𝑣𝑣 𝑡𝑡)/𝐻𝐻 2
[1.72]
Determining the field value of 𝐶𝐶𝑣𝑣 is difficult. Figure 1.24 provides a first-order determination of 𝐶𝐶𝑣𝑣 using the liquid limit (u. A. Department of the Navy, 1971). The value of ∆𝑢𝑢 for various depths (that is, 𝑧𝑧 = 0 to 𝑧𝑧 = 2𝐻𝐻) at time given time t (thus 𝑇𝑇𝑣𝑣 ) can be calculated from equation (71). The nature of this variation of ∆𝑢𝑢 is shown in figure 1.23b.
NPTEL - ADVANCED FOUNDATION ENGINEERING-1
Figure 1.24 Range of 𝐶𝐶𝑣𝑣 (after U. S. Department of the Navy, 1971) The average degree of consolidation of the clay layer can be defined as 𝑈𝑈 = 𝑆𝑆
𝑆𝑆𝑡𝑡
𝑚𝑚𝑚𝑚𝑚𝑚
[1.73]
Where
𝑈𝑈 = average degree of consolidation 𝑆𝑆𝑡𝑡 = settlement of a clay layer at time 𝑡𝑡 after the load application 𝑆𝑆𝑚𝑚𝑚𝑚𝑚𝑚 = maximum consolidation settlement that the clay will undergo under given loading If the initial pore water pressure (∆𝑢𝑢𝑜𝑜 ) distribution is constant with depth as shown in figure 1.25a, the average of consolidation can also be expressed as
NPTEL - ADVANCED FOUNDATION ENGINEERING-1
Figure 1.25 Drainage condition for consolidation: (a) two-way drainage; (b) oneway drainage
𝑈𝑈 = 𝑆𝑆
𝑆𝑆𝑡𝑡
𝑚𝑚𝑚𝑚𝑚𝑚
=
2𝐻𝐻
2𝐻𝐻
∫0 (∆𝑢𝑢 𝑜𝑜 )𝑑𝑑𝑑𝑑 −∫0 (∆𝑢𝑢)𝑑𝑑𝑑𝑑 2𝐻𝐻
∫0 (∆𝑢𝑢 𝑜𝑜 )𝑑𝑑𝑑𝑑
[1.74]
Or 𝑈𝑈 =
2𝐻𝐻
(∆𝑢𝑢 𝑜𝑜 )2𝐻𝐻−∫0 (∆𝑢𝑢)𝑑𝑑𝑑𝑑 (∆𝑢𝑢 𝑜𝑜 )2𝐻𝐻
=1−
2𝐻𝐻
∫0 (∆𝑢𝑢)𝑑𝑑𝑑𝑑 2𝐻𝐻(∆𝑢𝑢 𝑜𝑜 )
[1.75]
Now, combining equations (71 and 75) we obtain 𝑈𝑈 = 𝑆𝑆
𝑆𝑆𝑡𝑡
𝑚𝑚𝑚𝑚𝑚𝑚
2
=∞ −𝑀𝑀 = 1 − ∑𝑚𝑚 𝑚𝑚 =0 �𝑀𝑀 2 � 𝑒𝑒
2 𝑇𝑇 𝑣𝑣
[1.76]
The variation of 𝑈𝑈 with 𝑇𝑇𝑣𝑣 can be calculated from equation (76) and is plotted in figure 1.26. Note that equation (76) and thus figure 1.26 are also valid when an impermeable layer is located at the bottom of the clay layer (figure 1.25b). In that case, excess pore water pressure dissipation can take place in one direction only.
NPTEL - ADVANCED FOUNDATION ENGINEERING-1
Figure 1.26 Plot of time factor against average degree of consolidation (∆𝑢𝑢𝑜𝑜 = constant)
The length of the maximum drainage path then is equal to 𝐻𝐻 = 𝐻𝐻𝑐𝑐 .
The variation of 𝑇𝑇𝑣𝑣 with 𝑈𝑈 shown in figure 1.26 can also be approximated by 𝜋𝜋
𝑈𝑈% 2
𝑇𝑇𝑣𝑣 = 4 �100 �
(for 𝑈𝑈 = 0 − 60%)
[1.77]
𝑇𝑇𝑣𝑣 = 1.781 − 0.933 log(100 − 𝑈𝑈%) (for 𝑈𝑈 > 60%)
[1.78]
And
Sivaram and Swamee (1977) have also developed an empirical relationship between 𝑇𝑇𝑣𝑣 and 𝑈𝑈 that is valid for U varying from 0 to 100%. It is of the form 𝑇𝑇𝑣𝑣 =
𝜋𝜋 𝑈𝑈 % 2 � � 4 100 0.357 𝑈𝑈 % 5.6 �1−� � � 100
� �
[1.79]
In some cases, initial excess pore water pressure may not be constant with depth as shown in figure 1.25. Following are a few cases of those and the solutions for the average degree of consolidation.
NPTEL - ADVANCED FOUNDATION ENGINEERING-1 Trapezoidal Variation Figure 1.27 shows a trapezoidal variation of initial excess pore water pressure with two-way drainage. For this case the variation of 𝑇𝑇𝑣𝑣 with 𝑈𝑈 will be the same as shown in figure 1.26.
Figure 1.27 Trapezoidal initial excess pore water pressure distribution Sinusoidal Variation This variation is shown in figures 1.28a and 1.28b. For the initial excess pore water pressure variation shown in figure 1.28a,
z Figure 1.28 Sinusoidal initial excess pore water pressure distribution
NPTEL - ADVANCED FOUNDATION ENGINEERING-1 𝜋𝜋𝜋𝜋
∆𝑢𝑢 = ∆𝑢𝑢𝑜𝑜 𝑠𝑠𝑠𝑠𝑠𝑠 2𝐻𝐻
[1.80]
Similarly, for the case shown in figure 1.28b, 𝜋𝜋𝜋𝜋
∆𝑢𝑢 = ∆𝑢𝑢𝑜𝑜 𝑐𝑐𝑐𝑐𝑐𝑐 4𝐻𝐻
[1.81]
The variations of 𝑇𝑇𝑣𝑣 with 𝑈𝑈 for these two cases are shown in figure 1.29
Figure 1.29 Variation of 𝑈𝑈 with 𝑇𝑇𝑣𝑣 − sinusoidal variation of initial excess pore water pressure distribution Triangular Variation Figures 1.30 and 1.31 show several types of initial pore water pressure variation and the variation of 𝑇𝑇𝑣𝑣 with the average degree of consolidation.
NPTEL - ADVANCED FOUNDATION ENGINEERING-1
z Figure 1.30 Variation of 𝑈𝑈 with 𝑇𝑇𝑣𝑣 − triangular initial excess pore water pressure distribution
Figure 1.31 triangular initial excess pore water pressure distribution-variation of 𝑈𝑈 with 𝑇𝑇𝑣𝑣
NPTEL - ADVANCED FOUNDATION ENGINEERING-1
Example 9 A laboratory consolidation test on normally consolidated clay showed the following Load, 𝑝𝑝(kN/m2 )
Void ratio at the end of consolidation, e
140
0.92
212
0.86
The specimen tested was 25.4 mm in thickness and drained on both sides. The time required for the specimen to reach 50% consolidation was 4.5 min. A similar clay layer in the field, 2.8 m thick and drained on both sides, is subjected to similar average pressure increase (that is, 𝑝𝑝𝑜𝑜 = 140 kN/m2 and po + ∆p = 212kN/m2 ). Determine the a. Expected maximum consolidation settlement in the field b. Length of time required for the total settlement in the field to reach 40 mm (assume uniform initial excess pore water pressure increase with depth) Solution Part a For normally consolidated clay [equation 56] 𝐶𝐶𝑐𝑐 =
𝑒𝑒1 −𝑒𝑒2 𝑝𝑝 𝑙𝑙𝑙𝑙𝑙𝑙 � 2 � 𝑝𝑝 1
0.92−0.86
=
𝑙𝑙𝑙𝑙𝑙𝑙 �
From equation (64) 𝐶𝐶 𝐻𝐻
𝑐𝑐 𝑐𝑐 𝑆𝑆 = 1+𝑒𝑒 𝑙𝑙𝑙𝑙𝑙𝑙 𝑜𝑜
Part b
𝑝𝑝 𝑜𝑜 +∆𝑝𝑝 𝑝𝑝 𝑜𝑜
212 � 140
=
= 0.333
(0.333)(2.8) 1+0.92
212
𝑙𝑙𝑙𝑙𝑙𝑙 140 = 0.0875 m = 87.5 mm
From equation (73) the average degree of consolidation is 𝑈𝑈 = 𝑆𝑆
𝑆𝑆𝑡𝑡
𝑚𝑚𝑚𝑚𝑚𝑚
40
= 87.5 (100) = 45.7%
The coefficient of consolidation, 𝐶𝐶𝑣𝑣 , can be calculated from the laboratory test. From equation (72) 𝑇𝑇𝑣𝑣 =
𝐶𝐶𝑣𝑣 𝑡𝑡 𝐻𝐻 2
NPTEL - ADVANCED FOUNDATION ENGINEERING-1 For 50% consolidation (figure 1.26), 𝑇𝑇𝑣𝑣 = 0.197, 𝑡𝑡 = 4.5 min, and 𝐻𝐻 = 𝐻𝐻𝑐𝑐 /2 = 12.7 mm, so 𝐶𝐶𝑣𝑣 = 𝑇𝑇50
𝐻𝐻 2 𝑡𝑡
=
(0.197)(12.7)2 4.5
= 7.061 mm2 /min
Again, for field consolidation, 𝑈𝑈 = 45.7%. From equation (77) 𝜋𝜋
𝑈𝑈% 2
𝜋𝜋
45.7 2
𝑇𝑇𝑣𝑣 = 4 �100 � = 4 � 100 � = 0.164 But
𝑇𝑇𝑣𝑣 = Or
𝑡𝑡 =
𝐶𝐶𝑣𝑣 𝑡𝑡 𝐻𝐻 2
𝑇𝑇𝑣𝑣 𝐻𝐻 2 𝐶𝐶𝑣𝑣
=
0.164�
2.8×1000 2 � 2
7.061
= 45,523 min = 31.6 days
