Question

If two identical soil specimen were tested in a triaxial apparatus.

	${\sigma }_C\left(kPa\right)$	$\left({\sigma }_d{)}_f\right(kPa)$
Specimen 1	250	750
Specimen 2	500	1300

If the same soil is tested in direct shear test with a normal stress of 450 kN/m${}^2$, the shear stress at failure is

• A. $270kN/m^2$
• B. $333kN/m^2$
• C. $170kN/m^2$
• D. $450kN/m^2$

Answer 1

The correct option is B $333kN/m^2$

From plastic equilibrium equation
${\sigma }_1={\sigma }_3{\tan }^2{\alpha }_f+2c\tan {\alpha }_f$

$\begin{array}{cc}Specimen1& Specimen2\\ {\sigma }_3={\sigma }_c=250kPa& {\sigma }_3={\sigma }_c=500kPa\\ {\sigma }_1={\sigma }_c+{\sigma }_d=250+750& {\sigma }_1=500+1300\\ =1000kN/m^2& =1800kPa\end{array}$


$\Rightarrow {\tan }^2{\alpha }_f=3.2$

$\Rightarrow {\alpha }_f={60.79}^{\circ }$

$45+\varphi /2={60.79}^{\circ }$

$\varphi ={31.59}^{\circ }$

form (1)
$\begin{array}{cc}1000& =250\times {\tan }^2{60.79}^{\circ }+2c\times \tan {60.79}^{\circ }\\ 1000& =250\times 3.2+2c\times 1.79\\ 200& =2c\times 1.79\end{array}$

$\therefore c=55.87kPa$

form direct shear test:
$\sigma =450kN/m^2$
$\therefore {\tau }_f=s=c+\sigma .\tan \varphi$
$=55.87+450\times \tan {31.59}^{\circ }$
$=332.60kPa\approx 333kPa$