Question

A test was conducted on normally consolidated clay. Failure occurred under a allround pressure of $200kN/m^2$ and at deviatric stress of $450kN/m^2$. Match List-I with List-II and select the correct answer using codes below.

$\begin{array}{cc}\text{List-I}& \text{List-II}\\ \text{Angles made by failure plane with major principal axis}& \text{191}kN/m^2\\ \text{Angle made by plane of maximum shear stress with major principal plane}& \text{61}{}^{\circ }\\ \text{Shear stress on failure plane}& \text{305.6}kN/m^2\\ \text{Normal stress on failure plane}& \text{45}{}^{\circ }\end{array}$

Codes :

• A. P Q R S
  
  1 2 3 4
• B. P Q R S
  
  2 1 4 3
• C. P Q R S
  
  2 4 1 3
• D. P Q R S
  
  1 2 4 3

Answer 1

The correct option is C P Q R S


2 4 1 3
${\sigma }_c=200kN/m^2$

${\sigma }_d=450kN/m^2$

${\sigma }_3=200kN/m^2$

${\sigma }_1={\sigma }_c+{\sigma }_d$

$=650kN/m^2$

For $N.C.C\Rightarrow CD\Rightarrow C^{\prime }=0$

( A )
$c=0,\varphi =si{n}^{-1}\left(\frac{{\sigma }_1-{\sigma }_3}{{\sigma }_1+{\sigma }_3}\right)$

$\Rightarrow \varphi =si{n}^{-1}\left(\frac{650-200}{650+200}\right)={31.96}^{\circ }={32}^{\circ }$

${\alpha }_f=45+\frac{{32}^{\circ }}{2}$
${\alpha }_f={61}^{\circ }$

( B )
$\therefore$ Angle made by maximum shear stress is
$\alpha ={45}^{\circ }$


( C )
${\tau }_f=\frac{{\sigma }_1-{\sigma }_3}{2}sin2{\alpha }_f$

${\tau }_f=\frac{{\sigma }_d}{2}sin2{\alpha }_f=\frac{450}{2}sin(2\times {61}^{\circ })$

$\therefore {\tau }_f=191kN/m^2$


( D ) $\sigma =\frac{{\sigma }_1+{\sigma }_3}{2}+\frac{{\sigma }_1-{\sigma }_3}{2}cos2{\alpha }_f$

$=\sigma =\frac{650+200}{2}+\frac{650-200}{2}cos(2\times {61}^{\circ })$

$\therefore \sigma =305.6kN/m^2$