Question 1 |
The void ratio of a clay soil sample M decreased from 0.575 to 0.510 when the applied pressure is increased from 120 kPa to 180 kPa. For the same increment in pressure, the void ratio of another clay soil sample N decreases from 0.600 to 0.550. If the ratio of hydraulic conductivity of sample M to sample N is 0.125, then the ratio of coefficient of consolidation of sample M to sample N (round off to three decimal places) is ______________
0.124 | |
0.095 | |
0.002 | |
0.652 |
Question 1 Explanation:
\begin{aligned} m_{v}&=\frac{a_{V}}{1+e_{0}}=\frac{\Delta e}{\left(1+e_{0}\right) \times(\Delta \bar{\sigma})} & (\Delta \bar{\sigma} \text { is same for both } M \text { and } N)\\ m_{v 1}&=\frac{0.575-0.510}{(1+0.575) \times \Delta \bar{\sigma}}\\ m_{v 2}&=\frac{0.600-0.550}{(1+0.600) \times \Delta \bar{\sigma}}\\ \frac{C_{v_{1}}}{C_{V_{2}}} &=\frac{\frac{k_{1}}{m_{v_{1}} \gamma_{w}}}{\frac{k_{2}}{m_{v_{2}} \gamma_{w}}}=\frac{k_{1}}{k_{2}} \times \frac{m_{v_{2}}}{m_{v_{1}}} \\ &=0.125\left(\frac{1.575}{1.6}\right) \times\left(\frac{0.60-0.55}{0.575-0.510}\right) \\ &=0.0947 \simeq 0.095 \end{aligned}
Question 2 |
The soil profile at a road construction site is as shown in figure (not to scale). A large embankment is to be constructed at the site. The ground water table (GWT) is located at the surface of the clay layer, and the capillary rise in the sandy soil is negligible. The effective stress at the middle of the clay layer after the application of the embankment loading is 180 \mathrm{kN} / \mathrm{m}^{2}. Take unit weight of water, \gamma_{w}=9.81 \mathrm{kN} / \mathrm{m}^{3}.
The primary consolidation settlement (in m, round off to two decimal places) of the clay layer resulting from this loading will be ______
The primary consolidation settlement (in m, round off to two decimal places) of the clay layer resulting from this loading will be ______
0.12 | |
0.88 | |
0.45 | |
0.33 |
Question 2 Explanation:
Primary consolidation settlement
\Delta H=\frac{C_{c} H}{1+e_{0}} \log _{10}\left(\frac{\bar{\sigma}_{0}+\Delta \bar{\sigma}}{\bar{\sigma}_{0}}\right)
\bar{\sigma}_{0}+\Delta \bar{\sigma}=Effective stress at the center of clay layer after embankment loading
=180 \mathrm{kN} / \mathrm{m}^{2}
\bar{\sigma}_{0}= Effective stress at the centre of clay layer before embankment loading
\begin{aligned} \gamma_{\text {sub }} \text { of clay layer } &=\frac{G-1}{1+e} \gamma_{w}=\frac{(2.65-1)}{1+\frac{w G}{1}} \times 9.81 \\ &=\frac{1.65 \times 9.81}{1+0.45 \times 2.65}=7.383 \mathrm{kN} / \mathrm{m}^{3} \\ \bar{\sigma}_{0} &=(18.5 \times 2)+(7.383) \times 3=59.149 \mathrm{kNm} /{ } \\ e_{0} &=\frac{w G}{1}=0.45 \times 2.65=1.1925 \\ \Rightarrow \qquad \qquad \qquad \qquad \qquad & \Delta \mathrm{H}=\frac{0.25 \times 6}{1+1.1925} \log _{10}\left(\frac{180}{59.149}\right)=0.33 \mathrm{~m} \end{aligned}
\Delta H=\frac{C_{c} H}{1+e_{0}} \log _{10}\left(\frac{\bar{\sigma}_{0}+\Delta \bar{\sigma}}{\bar{\sigma}_{0}}\right)
\bar{\sigma}_{0}+\Delta \bar{\sigma}=Effective stress at the center of clay layer after embankment loading
=180 \mathrm{kN} / \mathrm{m}^{2}
\bar{\sigma}_{0}= Effective stress at the centre of clay layer before embankment loading
\begin{aligned} \gamma_{\text {sub }} \text { of clay layer } &=\frac{G-1}{1+e} \gamma_{w}=\frac{(2.65-1)}{1+\frac{w G}{1}} \times 9.81 \\ &=\frac{1.65 \times 9.81}{1+0.45 \times 2.65}=7.383 \mathrm{kN} / \mathrm{m}^{3} \\ \bar{\sigma}_{0} &=(18.5 \times 2)+(7.383) \times 3=59.149 \mathrm{kNm} /{ } \\ e_{0} &=\frac{w G}{1}=0.45 \times 2.65=1.1925 \\ \Rightarrow \qquad \qquad \qquad \qquad \qquad & \Delta \mathrm{H}=\frac{0.25 \times 6}{1+1.1925} \log _{10}\left(\frac{180}{59.149}\right)=0.33 \mathrm{~m} \end{aligned}
Question 3 |
A clay layer of thickness H has a preconsolidation pressure p_{c}
and an initial void ratio e_{0}.The initial effective overburden stress at the mid-height of the layer is p_{0}. At the same location, the increment in effective stress due to applied external load is \Delta p. The compression and swelling indices of the clay are C_{c}
and C_{s}, respectively. If p_{0} \lt p_{c} \lt \left(p_{0}+\Delta p\right), then the correct expression to estimate the consolidation settlement \left(S_{c}\right) of the clay layer is
s_{c}=\frac{H}{I+e_{0}}\left[C_{c} \log \frac{p_{c}}{p_{0}}+C_{s} \log \frac{p_{0}+\Delta p}{p_{c}}\right] | |
s_{c}=\frac{H}{I+e_{0}}\left[C_{s} \log \frac{p_{c}}{p_{0}}+C_{c} \log \frac{p_{0}+\Delta p}{p_{c}}\right] | |
s_{c}=\frac{H}{I+e_{0}}\left[C_{c} \log \frac{p_{0}}{p_{c}}+C_{s} \log \frac{p_{0}+\Delta p}{p_{c}}\right] | |
s_{c}=\frac{H}{I+e_{0}}\left[C_{s} \log \frac{p_{0}}{p_{c}}+C_{c} \log \frac{p_{0}+\Delta p}{p_{c}}\right] |
Question 3 Explanation:
As the soil is initially in over consolidate state (p_o \lt p_c), due to the external applied load (\Delta p) the soil initially undergoes re-compression upto (p_c) and then changes to virgin compression from (p_c) to (p_f)
The re-compression index (C_r) is also called Swelling index, (C_S).
s_{c}=\frac{H}{I+e_{0}} \left[C_{s} \log \frac{p_{c}}{p_{0}}+C_{c} \log \frac{p_{0}+\Delta p}{p_{c}}\right]
Question 4 |
The soil profile at a construction site is shown in the figure (not to scale). Ground water table (GWT) is at 5 m below the ground level at present. An old well data shows that the ground water table was as low as 10 m below the ground level in the past. Take unit weight of water, \gamma_{W}=9.81 \mathrm{kN} / \mathrm{m}^{3}.
The overconsolidation ratio (OCR)(round off to two decimal places) at the mid-point of the clay layer is _______
The overconsolidation ratio (OCR)(round off to two decimal places) at the mid-point of the clay layer is _______
1.85 | |
1.22 | |
2.14 | |
0.56 |
Question 4 Explanation:
\begin{aligned} \mathrm{OCR} &=\frac{\bar{\sigma}_{c}}{\bar{\sigma}_{0}} \\ \bar{\sigma}_{c} &=\text { Preconsolidation stress } \\ \bar{\sigma}_{0} &=\text { Present effective stress } \\ \bar{\sigma}_{c} &=17.5 \times 10+(18.5-9.81) \times 5+(17-9.81) \times 4 \\ &=247.21 \mathrm{kN} / \mathrm{m}^{2} \\ \bar{\sigma}_{0} &=17.5 \times 5+(18.5-9.81) \times 10+(17-9.81) \times 4 \\ &=203.16 \mathrm{kN} / \mathrm{m}^{2} \\ \mathrm{OCR} &=\frac{247.21}{203.16}=1.22 \end{aligned}
Question 5 |
A footing of size 2m x 2m transferring a pressure of 200 kN/m^2, is placed at a depth
of 1.5 m below the ground as shown in the figure (not drawn to the scale). The clay
stratum is normally consolidated. The clay has specific gravity of 2.65 and compression
index of 0.3.
Considering 2:1 (vertical to horizontal) method of load distribution and \gamma _w = 10 kN/m^3, the primary consolidation settlement (in mm, round off to two decimal places) of the clay stratum is _________.
Considering 2:1 (vertical to horizontal) method of load distribution and \gamma _w = 10 kN/m^3, the primary consolidation settlement (in mm, round off to two decimal places) of the clay stratum is _________.
45.25 | |
74.24 | |
88.46 | |
92.47 |
Question 5 Explanation:
For clay layer,
\begin{aligned} \gamma _{sat}&=\left ( \frac{G+e}{1+e} \right )\gamma _w \\ &= \left ( \frac{2.65+1}{1+e} \right )10=17kNm^3\\ e_0&=1.357 \\ H_0&=1.5m \\ (\bar{\sigma _0})_{c-c}&=[2\gamma _d+0.5\gamma _{sat}+0.75\gamma _{sat}]-1.25\gamma _w \\ &= (2 \times 15+0.5 \times 15+0.75 \times 17)-1.25 \times 10\\ &= 39.25kN/m^2\\ \Delta \bar{\sigma }&=\frac{Force}{Area} \\ &= \frac{qB^2}{(B+2nZ)^2}\\ &=\frac{200 \times 2 \times 2}{(2 \times 2 \times \frac{1}{2} \times 1.75)^2}\\ &=56.88kN/m^2\\ \Delta H&=\frac{H_0C_c}{1+e_0} \log \left ( \frac{\bar{\sigma _0}+\Delta \bar{\sigma }}{\bar{\sigma_0 }} \right )\\ &=\frac{1.5 \times 0.3}{1+1.357} \log \left ( \frac{39.25+56.88}{39.25} \right )\\ &=74.27mm \end{aligned}
Question 6 |
A one-dimensional consolidation test is carried out on a standard 19 mm thick clay
sample. The oedometer's deflection gauge indicates a reading of 2.1 mm, just
before removal of the load, without allowing any swelling. The void ratio is 0.62
at this stage. The initial void ratio (round off to two decimal places) of the standard
specimen is ________.
0.25 | |
0.38 | |
0.96 | |
0.82 |
Question 6 Explanation:
Oedometer reading = 2.1 mm
16 mm thick, e = 0.62
\begin{aligned} \frac{\Delta H}{H_0}&=\frac{\Delta e}{1+e_0}=\frac{e_0-e_f}{1+e_0} \\ \frac{2.1 mm}{19mm} &=\frac{e_0-0.62}{1+e_0} \\ e_0&=0.82 \end{aligned}
16 mm thick, e = 0.62
\begin{aligned} \frac{\Delta H}{H_0}&=\frac{\Delta e}{1+e_0}=\frac{e_0-e_f}{1+e_0} \\ \frac{2.1 mm}{19mm} &=\frac{e_0-0.62}{1+e_0} \\ e_0&=0.82 \end{aligned}
Question 7 |
A 10 m thick clay layer is resting over a 3 m thick sand layer and is submerged. A
fill of 2 m thick sand with unit weight of 20 kN/m^3 is placed above the clay layer to
accelerate the rate of consolidation of the clay layer. Coefficient of consolidation of clay
is 9 \times 10^{-2} m^2/year and coefficient of volume compressibility of clay is 2.2 \times 10^{-4} m^2/kN.
Assume Taylor?s relation between time factor and average degree of consolidation.
The settlement (in mm, round off to two decimal places) of the clay layer, 10 years after the construction of the fill, is _________.
The settlement (in mm, round off to two decimal places) of the clay layer, 10 years after the construction of the fill, is _________.
40.25 | |
12.26 | |
24.33 | |
18.83 |
Question 7 Explanation:
\begin{aligned} \Delta \bar{\sigma }&=2 \times 20=40 kN/m^2\\ \Delta H&=m_v\Delta \bar{\sigma }H \\ &= 2.2 \times 10^{-4} \times 40 \times 10 \times 10^3 mm\\ &= 88 mm\\ T_v&=\frac{C \times t}{H^2} \\ &= \frac{9 \times 10^{-2}\times 10}{5^2}=0.036\\ T_v&=\frac{\pi}{4}U^2 \\ U&=\sqrt{\frac{0.036 \times 4}{\pi}}=0.214 \end{aligned}
\Delta h after 10 year =0.214 \times 88=18.832 mm
\Delta h after 10 year =0.214 \times 88=18.832 mm
Question 8 |
Construction of a new building founded on a clayey soil was completed in January 2010. In January 2014, the average consolidation settlement of the foundation in clay was recorded as 10 mm. The ultimate consolidation settlement was estimated in design as 40 mm. Considering double drainage to occur at the clayey soil site, the expected consolidation settlement in January 2019 (in mm, round off to the nearestinteger) will be ________
10 | |
12 | |
15 | |
20 |
Question 8 Explanation:
Jan. 2010 - Jan. 2014 (4 years)
Settlement is 10 mm
\Delta H = 40 mm
2 - way drainage
Settlement in Jan. 2019 (in 9 years) = ?
In 4 years,
\begin{aligned} \%U &=\frac{\Delta h}{\Delta H}\times 100 \\ &= \frac{10}{40}\times 100=25\%\\ T_v&=C_v\frac{t}{d^2}\;\;\;\because \%U \lt 60\% \\ T_v&=\frac{\pi}{4}U^2 \\ \frac{\pi}{4}\left ( \frac{25}{100} \right )^2&=\frac{C_v}{d^2} \times 4 \; years...(1) \end{aligned}
In 9 years
\begin{aligned} T_v&=C_v\frac{t}{d^2}\\ &=\frac{\pi }{4} \times\frac{(0.25)^2}{4}\times 9 \; \text{years}\\ &=0.11044\\ \because \; (T_v)_{60}&=0.283 \\ \therefore \;\%U &\lt 60\%\\ \therefore \;T_v&=0.11044=\frac{\pi}{4}U^2\\ \%U&=0.3749=37.499\%\\ \%U&=\frac{\Delta h}{\Delta H} \times 100\\ &=\frac{\Delta h}{40mm}\times 100=37.499\\ \Delta h&=14.99mm=15mm \end{aligned}
Settlement is 10 mm
\Delta H = 40 mm
2 - way drainage
Settlement in Jan. 2019 (in 9 years) = ?
In 4 years,
\begin{aligned} \%U &=\frac{\Delta h}{\Delta H}\times 100 \\ &= \frac{10}{40}\times 100=25\%\\ T_v&=C_v\frac{t}{d^2}\;\;\;\because \%U \lt 60\% \\ T_v&=\frac{\pi}{4}U^2 \\ \frac{\pi}{4}\left ( \frac{25}{100} \right )^2&=\frac{C_v}{d^2} \times 4 \; years...(1) \end{aligned}
In 9 years
\begin{aligned} T_v&=C_v\frac{t}{d^2}\\ &=\frac{\pi }{4} \times\frac{(0.25)^2}{4}\times 9 \; \text{years}\\ &=0.11044\\ \because \; (T_v)_{60}&=0.283 \\ \therefore \;\%U &\lt 60\%\\ \therefore \;T_v&=0.11044=\frac{\pi}{4}U^2\\ \%U&=0.3749=37.499\%\\ \%U&=\frac{\Delta h}{\Delta H} \times 100\\ &=\frac{\Delta h}{40mm}\times 100=37.499\\ \Delta h&=14.99mm=15mm \end{aligned}
Question 9 |
The compression curve (void ratio, e vs. effective stress, {\sigma_{v}}' ) for a certain clayey soil is a straight line in a semi-logarithmic plot and it passes through the points (e = 1.2; {\sigma_{v}}' = 50 kPa) and (e = 0.6; {\sigma_{v}}' = 800 kPa). The compression index (up to two decimal places) of the soil is ______
0 | |
0.25 | |
0.5 | |
1 |
Question 9 Explanation:
Compression Index
C_{c}=\frac{\Delta e}{\log \left(\frac{\bar{\sigma}_{1}}{\bar{\sigma}_{0}}\right)}=\frac{1.2-0.6}{\log \left(\frac{800}{50}\right)}
C_{c}=\frac{0.6}{\log (16)}=\frac{0.6}{1.204}=0.4982
C_{c}=\frac{\Delta e}{\log \left(\frac{\bar{\sigma}_{1}}{\bar{\sigma}_{0}}\right)}=\frac{1.2-0.6}{\log \left(\frac{800}{50}\right)}
C_{c}=\frac{0.6}{\log (16)}=\frac{0.6}{1.204}=0.4982
Question 10 |
The void ratio of a soil is 0.55 at an effective normal stress of 140 kPa. The compression index of the soil is 0.25. In order to reduce the void ratio to 0.4, an increase in the magnitude of effective normal stress (in kPa, up to one decimal place) should be ______
210.8 | |
350.6 | |
417.3 | |
125.7 |
Question 10 Explanation:
\begin{aligned} \Delta H &=\frac{H_{0} \Delta e}{1+e_{0}}=\frac{H_{0} C_{c}}{1+e_{0}} \log \left(\frac{\sigma_{0}+\Delta \sigma}{\sigma_{0}}\right) \\ \Delta H &=H_{0}\left(\frac{0.55-0.40}{1+0.50}\right)=\frac{H_{0} \times 0.25}{1+0.50} \log \left(\frac{140+\Delta \sigma}{140}\right) \\ \frac{0.15}{0.25} &=\frac{3}{5}=\log \left(\frac{140+\Delta \sigma}{140}\right) \\ \Delta \sigma &=417.35 \mathrm{kPa} \end{aligned}
There are 10 questions to complete.