Question 1 |
A 5 m high vertical wall has a saturated clay backfill. The saturation unit weight and
cohesion of clay are 18kN/m^3 and 20 kPa, respectively. The angle of internal friction
of clay is zero. In order to prevent development of tension zone, the height of the wall
is required to be increased. Dry sand is used as backfill above the clay for the increased
portion of the wall. The unit weight and angle of internal friction of sand are 16kN/m^3
and 30^{\circ}, respectively. Assume that the back of the wall is smooth and top of the backfill
is horizontal. To prevent the development of tension zone, the minimum height (in m,
round off to one decimal place) by which the wall has to be raised, is __________.
1.5 | |
2.5 | |
6.2 | |
5.2 |
Question 1 Explanation:
To prevent tension crack,
\begin{aligned} q&=\frac{2c}{\sqrt{k_a}}=\frac{2 \times 20}{1}=40\\ q&=\gamma _d x=40\\ x&=\frac{40}{16}=2.5m \end{aligned}
Question 2 |
The soil profile at a site up to a depth of 10 m is shown in the figure (not drawn to the
scale). The soil is preloaded with a uniform surcharge (q) of 70 kN/m^2 at the ground
level. The water table is at a depth of 3 m below ground level. The soil unit weight of
the respective layers is shown in the figure. Consider unit weight of water as 9.81 kN/m^3
and assume that the surcharge (q) is applied instantaneously.
Immediately after preloading, the effective stresses (in kPa) at points P and Q respectively, are
Immediately after preloading, the effective stresses (in kPa) at points P and Q respectively, are
124 and 204 | |
36 and 90 | |
36 and 126 | |
54 and 95 |
Question 2 Explanation:
Surcharge (q=70kN/m^2) is applied instantaneously hence excess pore pressure (u_i=70kPa) is developed at point P and Q [GWT level is at level P]
At point P: Total stress
\sigma =q+3\gamma =70+3\times 18
Pore water pressure = Hydrostatics pore pressure + Excess pore pressure
=0+u_i=0+70=70kN/m^2
Effective stress, \bar{\sigma }=\sigma -u=54kPa
At point Q: Total stress, \sigma= q+3\gamma +4\gamma _{sat}=70+3\times 18 +4 \times 20
Pore pressure, u = Hydrostatics pore pressure + Excess pore pressure
= 4\gamma _{w}+u_i=4 \times 9.81 +70
Effective stress, \bar{\sigma }=\sigma -u=94.76kPa
Question 3 |
Which one of the following statements is NOT correct?
A clay deposit with a liquidity index greater than unity is in a state of plastic consistency. | |
The cohesion of normally consolidated clay is zero when tri-axial test is conducted
under consolidated undrained condition. | |
The ultimate bearing capacity of a strip foundation supported on the surface of sandy
soil increase in direct proportion to the width of footing. | |
In case of a point load, Boussinesq's equation predicts higher value of vertical stress
at a point directly beneath the load as compared to Westergaard's equation. |
Question 3 Explanation:
A clay deposit with liquidty index greater then 1, will be in liquid stage of consistency.
\because \;\;I_L=\frac{w_n-w_p}{w_l-w_p} \gt 1
\therefore \;\; w_n \gt w_L
\because \;\;I_L=\frac{w_n-w_p}{w_l-w_p} \gt 1
\therefore \;\; w_n \gt w_L
Question 4 |
A 2mx 4m rectangular footing has to carry a uniformly distributed load of 120 kPa. As per the 2:1 dispersion method of stress distribution, the increment in vertical stress (in kPa) at a depth of 2m below the footing is ________
120 | |
80 | |
40 | |
20 |
Question 4 Explanation:
The area of rectangular footing = 2 m x 4 m
q = 120 kPa
as 2:1 dispersion method of stress distribution
\begin{aligned} \Delta \bar{\sigma }&=\frac{q(B \times L)}{(B+2nZ)(L+2nZ)}\\ &=\frac{120 \times 2 \times 4}{(2+2 \times \frac{1}{2}\times 2)(4+2 \times \frac{1}{2}\times 2)}\\ \Delta \bar{\sigma }&=40kPa \end{aligned}
q = 120 kPa
as 2:1 dispersion method of stress distribution
\begin{aligned} \Delta \bar{\sigma }&=\frac{q(B \times L)}{(B+2nZ)(L+2nZ)}\\ &=\frac{120 \times 2 \times 4}{(2+2 \times \frac{1}{2}\times 2)(4+2 \times \frac{1}{2}\times 2)}\\ \Delta \bar{\sigma }&=40kPa \end{aligned}
Question 5 |
A concentrated load of 500 kN is applied on an elastic half space. The ratio of the increase in vertical normal stress at depths of 2m and 4m along the point of the loading, as per Boussinesq's theory, would be ___
2 | |
3 | |
4 | |
5 |
Question 5 Explanation:
As per Boussinesq's equation,
The vertical stress below the ground level at a depth, Z, vertically below the load,
\begin{aligned} \sigma _Z&=\frac{3}{2\pi}\frac{Q}{Z^2}\\ \sigma _Z &\propto \frac{1}{Z^2}\\ \therefore \;\; \frac{\sigma _{Z_1}}{\sigma _{Z_2}}&=\left [ \frac{Z_2}{Z_1} \right ]^2\\ &=\left [ \frac{4}{2} \right ]^2=4 \end{aligned}
The vertical stress below the ground level at a depth, Z, vertically below the load,
\begin{aligned} \sigma _Z&=\frac{3}{2\pi}\frac{Q}{Z^2}\\ \sigma _Z &\propto \frac{1}{Z^2}\\ \therefore \;\; \frac{\sigma _{Z_1}}{\sigma _{Z_2}}&=\left [ \frac{Z_2}{Z_1} \right ]^2\\ &=\left [ \frac{4}{2} \right ]^2=4 \end{aligned}
Question 6 |
Which one of the following statements is NOT correct?
When the water content of soil lies between its liquid limit and plastic limit, the soil is said to be in plastic state. | |
Boussinesq's theory is used for the analysis of stratified soil. | |
The inclination of stable slope in cohesive soil can be greater than its angle of internal friction. | |
For saturated dense fine sand, after applying overburden correction, if the Standard Penetration Test value exceeds 15, dilatancy correction is to be applied. |
Question 6 Explanation:
Boussinesq's assumed soil as isotropic hence not applicable for stratified soil.
Question 7 |
Consider a square-shaped area ABCD on the ground with its Centre at M as shown in the figure. Four concentrated vertical load of P=5000 kN are applied on this area, at each corner.
The vertical stress increment (in kPa, up to one decimal place) due to these loads according to the Boussinesq's equation, at a point 5 m right below M, is_____
The vertical stress increment (in kPa, up to one decimal place) due to these loads according to the Boussinesq's equation, at a point 5 m right below M, is_____
190.8 | |
95.6 | |
99.8 | |
47.4 |
Question 7 Explanation:
\begin{array}{l} P=5000 \mathrm{kN} \\ r=\sqrt{2^{2}+2^{2}}=2 \sqrt{2} \mathrm{m} \end{array}
According to Boussinesq's
\sigma_{z}=k \frac{Q}{z^{2}}=\frac{3}{2 \pi}\left(\frac{1}{1+\frac{r^{2}}{z^{2}}}\right)^{5 / 2} \frac{Q}{z^{2}}
\Rightarrow Vertical stress at 5m below point M
\begin{aligned} \sigma_{z} &=4 \times \frac{3}{2 \pi} \times\left\{\frac{1}{1+\left(\frac{2 \sqrt{2}}{5}\right)^{2}}\right\}^{5/2} \quad \times \frac{5000}{5^{2}} \\ &=190.8 \mathrm{kPa} \end{aligned}
Question 8 |
A uniformly distributed line load of 500 kN/m is acting on the ground surface. Based on Boussinesq's theory, the ratio of vertical stress at a depth 2 m to that at 4 m, right below the line of loading, is
0.25 | |
0.5 | |
2 | |
4 |
Question 8 Explanation:
q^{\prime}=500 \mathrm{kN} / \mathrm{m}
x=0 (Just below the line load)
\begin{aligned} \sigma_{z} &=\frac{2 q^{\prime}}{\pi z}\left(\frac{1}{1+\frac{x^{2}}{z^{2}}}\right)^{2}=\frac{2 q^{\prime}}{\pi z} \\ \text { Ratio } &=\frac{\left(\sigma_{z}\right)_{a t, 2 m}}{\left(\sigma_{z}\right)_{a t, 4 m}}=\frac{\frac{2 q^{\prime}}{\pi \times 2}}{\frac{2 q^{\prime}}{\pi \times 4}}=\frac{4}{2}=2 \end{aligned}
x=0 (Just below the line load)
\begin{aligned} \sigma_{z} &=\frac{2 q^{\prime}}{\pi z}\left(\frac{1}{1+\frac{x^{2}}{z^{2}}}\right)^{2}=\frac{2 q^{\prime}}{\pi z} \\ \text { Ratio } &=\frac{\left(\sigma_{z}\right)_{a t, 2 m}}{\left(\sigma_{z}\right)_{a t, 4 m}}=\frac{\frac{2 q^{\prime}}{\pi \times 2}}{\frac{2 q^{\prime}}{\pi \times 4}}=\frac{4}{2}=2 \end{aligned}
Question 9 |
The vertical stress at point P1 due to the point load Q on the ground surface as
shown in figure is \sigma _{2}. According to Boussinesq's equation, the vertical stress at point P2 shown in figure will be
\sigma _{z}/2 | |
\sigma _{z} | |
2\sigma _{z} | |
4\sigma _{z} |
Question 9 Explanation:
As per Boussinesq equation, the vertical stress \sigma_{2} at a point located at a depth 'z' and a horizontal distance 'r' from the point of application of point load Q is
\sigma_{z}=\frac{3 Q}{2 \pi z^{2}}\left[\frac{1}{1+\left(\frac{r}{z}\right)^{2}}\right]^{5 / 2}
For a horizontal distance \frac{r}{2} and a depth \frac{2}{2}, the stress will be
=\frac{3 Q}{2 \pi\left(\frac{z}{2}\right)^{2}}\left[\frac{1}{1+\left(\frac{r / 2}{2 / 2}\right)^{2}}\right]^{5 / 2}=4 \times \frac{3 Q}{2 \pi z^{2}}\left(\frac{1}{1+\left(\frac{r}{z}\right)^{2}}\right)^{5 / 2}
=4 \sigma_{z}
\sigma_{z}=\frac{3 Q}{2 \pi z^{2}}\left[\frac{1}{1+\left(\frac{r}{z}\right)^{2}}\right]^{5 / 2}
For a horizontal distance \frac{r}{2} and a depth \frac{2}{2}, the stress will be
=\frac{3 Q}{2 \pi\left(\frac{z}{2}\right)^{2}}\left[\frac{1}{1+\left(\frac{r / 2}{2 / 2}\right)^{2}}\right]^{5 / 2}=4 \times \frac{3 Q}{2 \pi z^{2}}\left(\frac{1}{1+\left(\frac{r}{z}\right)^{2}}\right)^{5 / 2}
=4 \sigma_{z}
Question 10 |
A footing 2m\times 1m
exerts a uniform pressure of 150 kN/m^{2} on the soil. Assuming a load dispersion of 2 vertical to 1 horizontal, the average vertical stress (kN/m^{2}) at 1.0 m below the footing is
50 | |
75 | |
80 | |
100 |
Question 10 Explanation:
Total dispersion area
\begin{aligned} &=(2+0.5+0.5) \times(1+0.5+0.5) \\ &=3 \mathrm{m} \times 2 \mathrm{m} \end{aligned}
There are 10 questions to complete.
Some of the questions are from another topic. So they have been placed here mistakenly.
I request you to please see into this.
Can you please write the question no
1,2,3