GATE CE 2006

The augmented matrix for given system is,
\left [ \left.\begin{matrix} 0 & 4 & 3\\ 2 & 0 & -1\\ 3 & 2 & 0 \end{matrix}\right|\begin{matrix} 8\\ 2\\ 5 \end{matrix} \right ]\xrightarrow[]{exchange 1st and 2nd row}\left [ \left.\begin{matrix} 2 &0 & -1\\ 0 & 4 & 3\\ 3 & 2 & 0 \end{matrix}\right|\begin{matrix} 2\\ 8\\ 5 \end{matrix} \right ]
then by Gauss elimination procedure
\left [ \left.\begin{matrix} 2 & 0 & -1\\ 0 & 4 & 3\\ 3 & 2 & 0 \end{matrix}\right|\begin{matrix} 2\\ 8\\ 5 \end{matrix} \right ]\xrightarrow[]{R_{3}-\frac{3}{2}R_{1}}\left [ \left.\begin{matrix} 2 &0 & -1\\ 0 & 4 & 3\\ 0 & 2 & 3/2 \end{matrix}\right|\begin{matrix} 2\\ 8\\ 2 \end{matrix} \right ]\xrightarrow[]{R_{3}-\frac{2}{4}R_{2}}\left [ \left.\begin{matrix} 2 & 0 & -1\\ 0 & 4 & 3\\ 0 & 0 & 0 \end{matrix}\right|\begin{matrix} 8\\ 8\\ -2 \end{matrix} \right ]
For last row we see 0=-2 which is inconsistent
\therefore Solution is non-existent for above system.

\frac{\mathrm{d} y}{\mathrm{d} x}=0.25y^{2} \;\;\left ( y=1\: at\: x=0 \right )
h=1
Iterative equation for backward (implicit) Euler methods for above equation would be,
\frac{y_{k+1}-y_{k}}{h}=0.25y^{2}_{k+1}
\Rightarrow \;\; y_{k+1}-y_{k}=0.25h y^{2}_{k+1}
\Rightarrow \;\; 0.25h y^{2}_{k+1}-y_{k+1}+y_{k}=0
putting k=0 in above equation
0.25h y^{2}_{1}-y_{1}+y_{0}=0
since y_{0}=1 and h=1
0.25 y_{1}^{2}-y_{1}+1=0
y_{1}=2

The maximum and minimum principal stresses are same. So, radius of circle becomes zero and centre is at (30, 0). The circle is represented by a point.

Cross-section of the shaft is the section perpendicular to the longitudinal axis of shaft. In case of twisting moment acting at its ends, normal stress is zero on the cross-section only shear stress acts.

The specific gravity of cohesionless granular soils (sands) does not vary much and for all practical purposes it is taken to be 2.65. Critical hydraulic gradient should be nearly 1 for quick sand condition.
i.e.\quad i_{cr}=\frac{G-1}{1+e}
From the above equation, the void ratio range is found to be between 0.6 and 0.7.

\begin{array}{l} O P=\text {Major stress}=\sigma_{1} \\ O Q=\text {Mirror stress}=\sigma_{3} \end{array}
Active condition:



OP = Vertical stress at base
=\sigma_{v} \cos \beta=\gamma+\cos \beta
OQ = Lateral earth pressure at base
=\mathrm{K}_{\mathrm{a}} \cdot \sigma_{\mathrm{v}} \cos \beta
Note: QR and QR' are failure plane for active case.

Capillary rise is more in fine grained soils and as a result of this, the capillary pressure is also more than the coarse grained soils.