Question 1 |
A saturated compressible clay layer of thickness h is sandwiched between two
sand layers, as shown in the figure. Initially, the total vertical stress and pore
water pressure at point P, which is located at the mid-depth of the clay layer,
were 150 kPa and 25 kPa, respectively. Construction of a building caused an
additional total vertical stress of 100 kPa at P. When the vertical effective stress
at P is 175 kPa, the percentage of consolidation in the clay layer at P is
_____________. (in integer)
25 | |
30 | |
50 | |
80 |
Question 1 Explanation:
Initial effective stress at P
\sigma _{P_1}'=150-25=125kPa
After construction,
\sigma _{P_2}'=125+100=225kPa
100% consolidation will occur when effective stress at P reaches 225 kPa.
Percentage of consolidation when vertical effective stress is at P is 175 kPa.
=\frac{175-125}{225-125} \times 100=\frac{50}{100} \times 100=50
\sigma _{P_1}'=150-25=125kPa
After construction,
\sigma _{P_2}'=125+100=225kPa
100% consolidation will occur when effective stress at P reaches 225 kPa.
Percentage of consolidation when vertical effective stress is at P is 175 kPa.
=\frac{175-125}{225-125} \times 100=\frac{50}{100} \times 100=50
Question 2 |
In a triaxial unconsolidated undrained (UU) test on a saturated clay sample, the
cell pressure was 100 kPa. If the deviatoric stress at failure was 150 kPa, then
the undrained shear strength of the soil is _________ kPa. (in integer)
25 | |
50 | |
75 | |
85 |
Question 2 Explanation:
\sigma _3=\sigma _c=100kPa , \sigma _d=\sigma _1-\sigma _3=150kPa
For UU test
Undrained shear strenth
C_{uu}=\tau _f=\frac{\sigma _1-\sigma _3}{2}=\frac{150}{2}=75KPa
For UU test
Undrained shear strenth
C_{uu}=\tau _f=\frac{\sigma _1-\sigma _3}{2}=\frac{150}{2}=75KPa
Question 3 |
A raft foundation of 30 m x 25 m is proposed to be constructed at a depth of 8 m
in a sand layer. A 25 m thick saturated clay layer exists 2 m below the base of
the raft foundation. Below the clay layer, a dense sand layer exists at the site.
A 25 mm thick undisturbed sample was collected from the mid-depth of the clay
layer and tested in a laboratory oedometer under double drainage condition. It
was found that the soil sample had undergone 50% consolidation settlement in
10 minutes.
The time (in days) required for 25% consolidation settlement of the raft foundation will be ______. (round off to the nearest integer)
The time (in days) required for 25% consolidation settlement of the raft foundation will be ______. (round off to the nearest integer)
3525 | |
1254 | |
1736 | |
2463 |
Question 3 Explanation:
\begin{aligned} (T_v)_{50}&=C_v\frac{t}{d^2} \;\;\; (\text{from lab})\\ \frac{\pi}{4}(0.5)^2&=C_v \times \frac{10 min}{\left ( \frac{25}{2} \times 10^{-3} \right )^2}\\ (T_v)_{25}&=C_v\frac{t}{d^2} \;\;\; (\text{from field})\\ \frac{\pi}{4}(0.25)^2&=C_v \frac{t}{(12.5)^2}\\ \end{aligned}
Since the soil is same -> C_v same
\begin{aligned} \frac{\pi}{4}(0.25)^2&= \frac{\frac{\pi}{4}(0.5)^2 \times (12.5 \times 10^{-3})^2}{10} \times \frac{t}{(12.5)^2}\\ t&=1736 \;days \end{aligned}
Question 4 |
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 4 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 5 |
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 5 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 6 |
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 6 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 7 |
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 7 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 8 |
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 8 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 9 |
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 9 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 10 |
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 10 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
There are 10 questions to complete.