Question

2. A consolidated undrained triaxial test, in which pore pressure measurements were made, was carried out on a saturated soil with the following results. Given that the initial diameter of the sample was 38 mm in all three cases, determine the values of cohesion and angle of internal friction with respect to (a) total stresses (c and o). (b) effective stresses (c' and o'). The minor and major principal stresses are calculated and presented in the table below (You are responsible of learning this procedure for the MT exam). Total Stress Cireles Effective Stress Circles Cell pressure o. (kN/m) Piston load P (kN) Pore maxial pressure, strain (kN/m') . A (m) loo) -P/A (kPa) la, 0:1/2 .0.1/2 lo' loa ',1/2 10.0 /2 kPa) (kPa) jo' o* 1/2 kPa kPa 70 139 184 139 0.35 0.54 0.81 24.5 91 189 10 0.0013 10 0.0013 100.0013 278 429 348 46 639 119 1063 231 323 548 874 209 424 242 214 333 214 420

Answer 1

Consolidated Undrained Tri-axial test is being conducted on the soil and the following observations are obtained:

Cell Pressure ($\sigma _{3}$) (KN/m2)	Deviator Stress ($\sigma _{d}=\sigma _{1}-\sigma _{3}$) (KN/m2)	Major Principal Stress ($\sigma _{1}$) (KN/m2)	Pore Pressure (u) (KN/m2)	Effective Cell Pressure ($\sigma _{3}'$) (KN/m2)	Effective Major Principal Stress ($\sigma _{1}'$) (KN/m2)
70	278	348	24.5	46	323
210	429	639	91	119	548
420	643	1063	189	231	874

We have to determine the values of cohesion and internal angle of friction.

This can easily be calculated using the tri-axial test equation, which is given by:

• (a). Corresponding to Total Stresses:

$\sigma _{1} = \sigma _{3}\tan ^{2}\left ( 45+\frac{\Phi }{2} \right )+2C.\tan \left ( 45+\frac{\Phi }{2} \right )$

On taking 1st and 2nd case:

$348 = 70.\tan ^{2}\left ( 45+\frac{\Phi }{2} \right )+2C.\tan \left ( 45+\frac{\Phi }{2} \right )$   ............(i)

$639 = 210.\tan ^{2}\left ( 45+\frac{\Phi }{2} \right )+2C.\tan \left ( 45+\frac{\Phi }{2} \right )$ ............(ii)

On solving equation (i) and (ii);

Subtract equation (i) from the equation (ii);

$291 = 140.\tan ^{2}\left ( 45+\frac{\Phi }{2} \right )$

and we got from here $\Phi = 20.5086^{\circ}$

on putting this value of $\Phi$ in equation (i);

$348 = 70.\tan ^{2}\left ( 45+\frac{20.5086 }{2} \right )+2C.\tan \left ( 45+\frac{20.5086 }{2} \right )$

We get the following results:

Cohesion, $\mathbf{C}$ = 70.23 KN/m2

The angle of Internal Friction, $\mathbf{\Phi = 20.5086^{\circ}}$

• (b). Corresponding to Effective Stresses:

$\sigma _{1}' = \sigma _{3}'.\tan ^{2}\left ( 45+\frac{\Phi' }{2} \right )+2C'.\tan \left ( 45+\frac{\Phi' }{2} \right )$

On taking 1st and 2nd case:

$323 = 46.\tan ^{2}\left ( 45+\frac{\Phi' }{2} \right )+2C'.\tan \left ( 45+\frac{\Phi' }{2} \right )$   ............(iii)

$548 = 119.\tan ^{2}\left ( 45+\frac{\Phi' }{2} \right )+2C'.\tan \left ( 45+\frac{\Phi' }{2} \right )$ ............(iv)

Similar to part (a) of the question,  solving equation (iii) and (iv):

We get the following result;

Cohesion, $\mathbf{C'}$ = 51.612 KN/m2

The angle of Internal Friction, $\mathbf{\Phi' = 30.668^{\circ}}$