Question 1 |
In the given figure, Point \mathrm{O} indicates the stress point of a soil element at initial non-hydrostatic stress condition. For the stress path (OP), which of the following loading conditions is correct ?
\sigma_{v} is increasing and \sigma_{h} is constant. | |
\sigma_{v} is constant and \sigma_{h} is increasing. | |
\sigma_{v} is increasing and \sigma_{h} is decreasing. | |
\sigma_{v} is decreasing and \sigma_{h} is increasing. |
Question 1 Explanation:
For 1: 1 slope, \frac{d q}{d P}=1
\Rightarrow \frac{d \sigma_{V}-d \sigma_{H}}{d \sigma_{V}+d \sigma_{H}}=1
\Rightarrow d \sigma_{V}-d \sigma_{H}=d \sigma_{V}+d \sigma_{H}
\Rightarrow \quad-\mathrm{d} \sigma_{\mathrm{H}}=\mathrm{d} \sigma_{\mathrm{H}}
\Rightarrow \quad \mathrm{d} \sigma_{\mathrm{H}}=0
Hence, if \sigma_{H} is constant then increasing \sigma_{V} or decreasing \sigma_{\mathrm{V}} will lead to 1: 1 slope.
Answer is (A)
Question 2 |
An unconfined compression strength test was conducted on a cohesive soil. The test specimen failed at an axial stress of 76 \mathrm{kPa}. The undrained cohesion (in \mathrm{kPa}, in integer) of the soil is ___
18 | |
38 | |
28 | |
54 |
Question 2 Explanation:
UCS Test :
\sigma_{3}=0 [Cell pressure]
\therefore Axial stress = Deviatoric stress =\sigma_{1}-\sigma_{3}=\sigma_{1}
Given, \sigma_{1}=76 \mathrm{kPa}
We know,
\sigma_{1}=\sigma_{3} \tan ^{2}\left(45^{\circ}+\frac{\phi}{2}\right)+2 C \tan \left(45^{\circ}+\frac{\phi}{2}\right)
[\mathrm{C}= Undrained cohesion, \phi=0 (Cohesive soil)]
\Rightarrow \quad \sigma_{1}=2 \mathrm{C}
\Rightarrow \quad 76=2 \mathrm{C}
\Rightarrow \quad C=38 \mathrm{kPa}
\sigma_{3}=0 [Cell pressure]
\therefore Axial stress = Deviatoric stress =\sigma_{1}-\sigma_{3}=\sigma_{1}
Given, \sigma_{1}=76 \mathrm{kPa}
We know,
\sigma_{1}=\sigma_{3} \tan ^{2}\left(45^{\circ}+\frac{\phi}{2}\right)+2 C \tan \left(45^{\circ}+\frac{\phi}{2}\right)
[\mathrm{C}= Undrained cohesion, \phi=0 (Cohesive soil)]
\Rightarrow \quad \sigma_{1}=2 \mathrm{C}
\Rightarrow \quad 76=2 \mathrm{C}
\Rightarrow \quad C=38 \mathrm{kPa}
Question 3 |
Consider that a force P is acting on the surface of a half-space (Boussingesq's problem). The expression for the vertical stress \left(\sigma_{z}\right) at any point (r, z) with the half-space is given as,
\sigma_{z}=\frac{3 P}{2 \pi} \frac{z^{3}}{\left(r^{2}+z^{2}\right)^{5 / 2}}
where, r is the radial distance, and z is the depth with downward direction taken as positive. At any given r, there is a variation of \sigma_{z} along z, and at a specific z, the value of \sigma_{z} will be maximum. What is the locus of the maximum \sigma_{z} ?
\sigma_{z}=\frac{3 P}{2 \pi} \frac{z^{3}}{\left(r^{2}+z^{2}\right)^{5 / 2}}
where, r is the radial distance, and z is the depth with downward direction taken as positive. At any given r, there is a variation of \sigma_{z} along z, and at a specific z, the value of \sigma_{z} will be maximum. What is the locus of the maximum \sigma_{z} ?
z^2=\frac{3}{2}r^2 | |
z^3=\frac{3}{2}r^2 | |
z^2=\frac{5}{2}r^2 | |
z^3=\frac{5}{2}r^2 |
Question 3 Explanation:
The expression for the vertical stress \left(\sigma_{2}\right) at any point (r, z) with the half-space is given as
\sigma_{z}=\frac{3 P}{2 \pi} \frac{z^{3}}{\left(r^{2}+z^{2}\right)^{5 / 2}}
For \sigma_{\mathrm{z}} to be maximum
\begin{aligned} & \frac{\sigma_{z}}{\partial z}=0 \Rightarrow \frac{3 P}{2 \pi} \frac{\partial}{\partial z}\left[\frac{z^{3}}{\left(r^{2}+z^{2}\right)^{5 / 2}}\right] \\ &=\frac{3 P}{2 \pi} \frac{\left[\left(r^{2}+z^{2}\right)^{5 / 2}-z^{3} \times \frac{5}{2}\left(r^{2}+z^{2}\right)^{3 / 2} \times 2 z\right]}{\left(r^{2}+z^{2}\right)^{5}}=0 \\ &=\left[\left(r^{2}+z^{2}\right)\left(3 z^{2}\right)-5 z^{4}\right]=0 \\ &=3 r^{2}+3 z^{2}=5 z^{2} \\ & z^{2}=\frac{3 r^{2}}{2} \end{aligned}
\sigma_{z}=\frac{3 P}{2 \pi} \frac{z^{3}}{\left(r^{2}+z^{2}\right)^{5 / 2}}
For \sigma_{\mathrm{z}} to be maximum
\begin{aligned} & \frac{\sigma_{z}}{\partial z}=0 \Rightarrow \frac{3 P}{2 \pi} \frac{\partial}{\partial z}\left[\frac{z^{3}}{\left(r^{2}+z^{2}\right)^{5 / 2}}\right] \\ &=\frac{3 P}{2 \pi} \frac{\left[\left(r^{2}+z^{2}\right)^{5 / 2}-z^{3} \times \frac{5}{2}\left(r^{2}+z^{2}\right)^{3 / 2} \times 2 z\right]}{\left(r^{2}+z^{2}\right)^{5}}=0 \\ &=\left[\left(r^{2}+z^{2}\right)\left(3 z^{2}\right)-5 z^{4}\right]=0 \\ &=3 r^{2}+3 z^{2}=5 z^{2} \\ & z^{2}=\frac{3 r^{2}}{2} \end{aligned}
Question 4 |
A concentrically loaded isolated square footing of size 2 m x 2 m carries a
concentrated vertical load of 1000 kN. Considering Boussinesq's theory of stress
distribution, the maximum depth (in m) of the pressure bulb corresponding to
10% of the vertical load intensity will be ________. (round off to two
decimal places)
1.25 | |
4.35 | |
5.36 | |
6.25 |
Question 4 Explanation:
Considering Boussinesq's theory of stress
distribution
\sigma _x=\frac{3Q}{2\pi z^2}\left [ \frac{1}{1+\left ( \frac{r}{z} \right )^2} \right ]^{5/2}
For r=0, \sigma _z=\frac{3Q}{2\pi z^2}
\sigma _z=0.1q=0.1 \times \frac{Q}{B^2}=\frac{0.1}{2^2}Q
\frac{1}{40}Q=\frac{3Q}{2 \pi z^2}
z^2=\frac{3 \times 40}{2 \pi}\Rightarrow z=4.37m
\sigma _x=\frac{3Q}{2\pi z^2}\left [ \frac{1}{1+\left ( \frac{r}{z} \right )^2} \right ]^{5/2}
For r=0, \sigma _z=\frac{3Q}{2\pi z^2}
\sigma _z=0.1q=0.1 \times \frac{Q}{B^2}=\frac{0.1}{2^2}Q
\frac{1}{40}Q=\frac{3Q}{2 \pi z^2}
z^2=\frac{3 \times 40}{2 \pi}\Rightarrow z=4.37m
Question 5 |
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 5 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}
There are 5 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
Please put figure for question number 6