Holooly Plus Logo
Holooly Plus Logo

Question 7.3: A stratum of normally consolidated clay 7 m thick is located...

A stratum of normally consolidated clay 7 m thick is located at a depth 12m below ground level. The natural moisture content of the clay is 40.5 per cent and its liquid limit is 48 per cent. The specific gravity of the solid particles is 2.76. The water table is located at a depth 5 m below ground surface. The soil is sand above the clay stratum. The submerged unit weight of the sand is 11 \mathrm{kN} / \mathrm{m}^{3} and the same weighs18 \mathrm{kN} / \mathrm{m}^{3} above the water table. The average increase in pressure at the center of the clay stratum is120 \mathrm{kN} / \mathrm{m}^{2} due to the weight of a building that will be constructed on the sand above the clay stratum. Estimate the expected settlement of the structure.

Question Data is a breakdown of the data given in the question above.
  • Thickness of the clay stratum: 7 m
  • Depth of the clay stratum below ground level: 12 m
  • Natural moisture content of the clay: 40.5%
  • Liquid limit of the clay: 48%
  • Specific gravity of the solid particles in the clay: 2.76
  • Depth of the water table below ground surface: 5 m
  • Soil above the clay stratum: sand
  • Submerged unit weight of the sand: 11 kN/m^3
  • Unit weight of the sand above the water table: 18 kN/m^3
  • Average increase in pressure at the center of the clay stratum due to the weight of the building: 120 kN/m^2.
The Blue Check Mark means that this solution has been answered and checked by an expert. This guarantees that the final answer is accurate.
Learn more on how we answer questions.
Step 1:
The first step is to determine the water content ratio (Ww/Ws) and the weight of solids (Ws) for the clay. The water content ratio is given as Wn = Ww/Ws, where Ww is the weight of water and Ws is the weight of solids. The weight of solids is calculated by multiplying the volume of solids (Vs) with the specific gravity of solids (Gs) and the unit weight of water (γw). In this case, Ws = Vs Gs γw = 1 × 2.76 × 1 = 2.76 g.
Step 2:
Next, we can calculate the weight of water (Ww) using the water content ratio and the weight of solids. Ww = (40.5/100), Ws = (40.5/100) 2.76 = 1.118 g.
Step 3:
The volume of water (Vw) can be calculated by dividing the weight of water by the unit weight of water. Vw = Ww/γw = 1.118/1 = 1.118 cm^3.
Step 4:
The initial void ratio (e0) can be determined by dividing the volume of water by the total volume (V) of the soil. In this case, e0 = Vw/V = 1.118/1 = 1.118.
Step 5:
The total weight (W) of the soil can be calculated by adding the weight of water and the weight of solids. W = Ww + Ws = 1.118 + 2.76 = 3.878 g.
Step 6:
The total unit weight (γt) of the soil can be determined by dividing the weight of the soil by the total volume (V) of the soil plus the volume of water (Vw). γt = W/(1 + e0) = 3.878/(1 + 1.118) = 1.83 g/cm^3.
Step 7:
The buoyant unit weight (γb) of the soil can be calculated by subtracting 1 (unit weight of water) from the total unit weight (γt). γb = γt - 1 = 1.83 - 1 = 0.83 g/cm^3.
Step 8:
The overburden pressure (p0) can be determined by multiplying the unit weights of the different layers of soil (γ1, γ2, γ3) with their respective heights (h1, h2, h3) and summing them up. In this case, p0 = γ1 h1 + γ2 h2 + γ3 h3 = 0.83 × 9.81 × 3.5 + 11 7 + 18 * 5 = 195.5 kN/m^2.
Step 9:
The compression index (Cc) can be calculated using the liquid limit (wl) of the soil. Cc = 0.009 (wl - 10) = 0.009 (48 - 10) = 0.34.
Step 10:
The excess pressure (Δp) is given as 120 kN/m^2.
Step 11:
The total settlement (St) can be determined using the compression index, initial void ratio (e0), overburden pressure (p0), and excess pressure (Δp). St = (Cc / (1 + e0)) H log((p0 + Δp) / p0) = (0.34 / 1.118) 700 log((195.5 + 120) / 195.5) = 23.3 cm.
Step 12:
Therefore, the estimated settlement is 23.3 cm.

Final Answer

1 . Determination of e and \gamma_{b} for the clay

 

\frac{W_{w}}{W_{s}}=w_{n}, W_{s}=V_{s} G_{s} \gamma_{w}=1 \times 2.76 \times 1=2.76 \mathrm{~g}

 

W_{w}=\frac{40.5}{100} \times 2.76=1.118 \mathrm{~g}

 

V_{w}=\frac{W_{w}}{\gamma_{w}}=\frac{1.118}{1}=1.118 \mathrm{~cm}^{3}

 

e_{0}=\frac{V_{w}}{V}=\frac{1.118}{1}=1.118

 

W=W_{w}+W_{s}=1.118+2.76=3.878 \mathrm{~g}

 

\gamma_{t}=\frac{W}{1+e_{0}}=\frac{3.88}{2.118}=1.83 \mathrm{~g} / \mathrm{cm}^{3}

 

\gamma_{b}=(1.83-1)=0.83 \mathrm{~g} / \mathrm{cm}^{3}

 

2. Determination of overburden pressure p_{0}

 

p_{0}=\gamma_{1} h_{1}+\gamma_{2} h_{2}+\gamma_{3} h_{3} \text { or }

 

p_{0}=0.83 \times 9.81 \times 3.5+11 \times 7+18 \times 5=195.5 \mathrm{kN} / \mathrm{m}^{2}

 

Compression index  \overline{P_{a}}=P_{a}+P_{w}=\frac{1}{2} \gamma_{b} H^{2} K_{A}+\frac{1}{2} \gamma_{w} H^{2} (11.17)

 

C_{c}=0.009\left(w_{l}-10\right)=0.009 \times(48-10)=0.34

 

4. Excess pressure

 

\Delta p=120 \mathrm{kN} / \mathrm{m}^{2}

 

5. Total Settlement

 

S_{t}=\frac{C_{c}}{1+e_{0}} H \log \frac{p_{0}+\Delta p}{p_{0}}

 

=\frac{0.34}{2118} \times 700 \log \frac{195.5+120}{1055}=23.3 \mathrm{~cm}

 

Estimated settlement = 23.3 cm.

7.3

Related Answered Questions

The loading period for a new building extended from May 1995 to May 1997. In May 2000, the average measured settlement was found to be 11.43 cm. It is known that the ultimate settlement will be about 35.56 cm. Estimate the settlement in May 2005. Assume double drainage to occur.

Verified Answer:
For the majority of practical cases in which loadi...

A strata of normally consolidated clay of thickness 10 ft is drained on one side only. It has a hydraulic conductivity of k = 1.863 x 10^-8 in/sec and a coefficient of volume compressibility mv = 8.6 x 10^-4 in^2/lb. Determine the ultimate value of the compression of the stratum by assuminga

Verified Answer:
Total compression,   S_{t}=m_{v} H \De...

The void ratio of a clay sample A decreased from 0.572 to 0.505 under a change in pressure from 122 to 180 kN/m^2. The void ratio of another sample B decreased from 0.61 to 0.557 under the same increment of pressure.The thickness of sample A was 1.5 times that of B. Nevertheless the time taken for

Verified Answer:
Let H_{a}= thickness of sample A, [...

A 2.5 cm thick sample of clay was taken from the field for predicting the time of settlement for a proposed building which exerts a uniform pressure of 100 kN/m^2 over the clay stratum. The sample was loaded to 100 kN/m^2 and proper drainage was allowed from top and bottom. It was seen that 50

Verified Answer:
T for 50% consolidation = 0.197.   The lab sa...

Two points on a curve for a normally consolidated clay have the following coordinates. Point 1 : e1, = 0.7, Pl = 2089 lb/ft^2 Point 2: e2 = 0.6, p2 = 6266 lb/ft^2 If the average overburden pressure on a 20 ft thick clay layer is 3133 lb/ft^2, how much settlement will the clay layer experience due

Verified Answer:
C_{c}=\frac{e_{0}-e}{\log p-\log p_{0}}=\f...

A soil sample has a compression index of 0.3. If the void ratio e at a stress of 2940 Ib/ft^2 is 0.5,compute (i) the void ratio if the stress is increased to 4200 Ib/ft^2, and (ii) the settlement of a soil stratum 13 ft thick.

Verified Answer:
Given:  C_{c}=0.3, e_{1}=0.50, p_{1}=2940 \...

Soil investigation at a site gave the following information. Fine sand exists to a depth of 10.6 m and below this lies a soft clay layer 7.60 m thick. The water table is at 4.60 m below the ground surface.The submerged unit weight of sand yb is 10.4 kN/m^3, and the wet unit weight above the water

Verified Answer:
For calculating settlement  S_{t}=\frac{C_{...

A column of a building carries a load of 1000 kips. The load is transferred to sub soil through a square footing of size 16 x 16 ft founded at a depth of 6.5 ft below ground level. The soil below the footing is fine sand up to a depth of 16.5 ft and below this is a soft compressible clay of

Verified Answer:
1. Required  \Delta p at the middle...

A recently completed fill was 32.8 ft thick and its initial average void ratio was 1.0. The fill was loaded on the surface by constructing an embankment covering a large area of the fill. Some months after the embankment was constructed, measurements of the fill indicated an average void ratio of

Verified Answer:
Per Eq. (7.7), the compression of the fill may be ...

During a consolidation test, a sample of fully saturated clay 3 cm thick (= h0) is consolidated under a pressure increment of 200 kN/m^2. When equilibrium is reached, the sample thickness is reduced to 2.60 cm. The pressure is then removed and the sample is allowed to expand and absorb water. The

Verified Answer:
Use equation (7.3)   \Delta e=\frac{1+...