Shear Strength of Soil

Mohr's failure theory

Mohr's theory states that a material fails when the shear stress on a plane reaches a limiting value that depends on the normal stress on that plane. Failure is therefore governed by a combination of normal and shear stress, not by the maximum stress alone:

$$\tau_f = f(\sigma)$$

The curve of $\tau_f$ against $\sigma$ is the failure envelope. Coulomb approximated it for soils by a straight line, giving the Mohr-Coulomb criterion:

$$\tau_f = c + \sigma\tan\phi \qquad \text{(effective stress: } \tau_f = c' + \sigma'\tan\phi'\text{)}$$

Here $c$ is cohesion and $\phi$ is the angle of internal friction. A stress state is safe if its Mohr circle lies below the line and fails when the circle touches it.

Derivation of the relation between $\sigma_1$ and $\sigma_3$

  tau |          . failure line
      |        .   tau = c + sigma tan(phi)
      |   _.-'-._.
      | .'   |   '.
      |/  phi|  C   \
 -----+-----+-+---+---+------ sigma
      c cot(phi) O   (s3)  (s1)

Consider the Mohr circle at failure, touching the failure line at point $T$. Its centre is $C$ at $\frac{\sigma_1+\sigma_3}{2}$ and radius is $\frac{\sigma_1-\sigma_3}{2}$. The failure line meets the $\sigma$-axis at $O'$, a distance $c\cot\phi$ to the left of the origin. In the right-angled triangle $O'TC$, the angle at $O'$ is $\phi$ and $CT \perp O'T$:

$$\sin\phi = \frac{CT}{O'C} = \frac{\dfrac{\sigma_1-\sigma_3}{2}}{c\cot\phi + \dfrac{\sigma_1+\sigma_3}{2}}$$ $$\sigma_1 - \sigma_3 = (\sigma_1+\sigma_3)\sin\phi + 2c\cos\phi$$ $$\sigma_1(1-\sin\phi) = \sigma_3(1+\sin\phi) + 2c\cos\phi$$ $$\boxed{\sigma_1 = \sigma_3\,\frac{1+\sin\phi}{1-\sin\phi} + 2c\,\frac{\cos\phi}{1-\sin\phi}}$$

Using $\frac{1+\sin\phi}{1-\sin\phi} = \tan^2\left(45^\circ+\frac{\phi}{2}\right)$ and $\frac{\cos\phi}{1-\sin\phi} = \tan\left(45^\circ+\frac{\phi}{2}\right)$:

$$\sigma_1 = \sigma_3\tan^2\left(45^\circ+\frac{\phi}{2}\right) + 2c\tan\left(45^\circ+\frac{\phi}{2}\right)$$

Equivalent form: $\sigma_3 = \sigma_1\tan^2\left(45^\circ-\frac{\phi}{2}\right) - 2c\tan\left(45^\circ-\frac{\phi}{2}\right)$.

Failure plane angle: the angle $\angle TCO'$ is $90^\circ+\phi$, so the failure plane is inclined at $\theta_f = 45^\circ + \frac{\phi}{2}$ to the major principal plane.

For cohesionless soil ($c=0$): $\sigma_1 = \sigma_3\tan^2(45^\circ+\phi/2)$. For a saturated clay in undrained loading ($\phi_u=0$): $\sigma_1 - \sigma_3 = 2c_u$.

Laboratory tests

1. Direct shear test (shear box).
2. Triaxial compression test: unconsolidated undrained (UU), consolidated undrained (CU) with or without pore pressure measurement, and consolidated drained (CD).
3. Unconfined compression test (UCS).
4. Laboratory vane shear test.
5. Ring shear test (for residual strength).

Field tests

1. Field (in-situ) vane shear test.
2. Standard Penetration Test (SPT).
3. Cone Penetration Test (CPT / static cone).
4. Pressuremeter test.
5. Plate load test.

Suitable tests for each soil

Mohr-Coulomb failure criterion: a soil fails when the shear stress on any plane reaches the shear strength, which is a linear function of the normal stress on that plane:

$$\tau_f = c + \sigma\tan\phi$$

In terms of effective stress, $\tau_f = c' + \sigma'\tan\phi'$. Here $c$ is the cohesion intercept and $\phi$ is the angle of shearing resistance (internal friction).

• If the Mohr circle for a stress state lies wholly below the failure line, the soil is stable.
• If the circle touches the line, the soil is on the point of failure.
• A circle cannot extend above the line.

Failure occurs on a plane inclined at $45^\circ+\phi/2$ to the major principal plane.

Mohr-Coulomb theory: the shear strength of a soil on a plane is a linear function of the normal stress on it, $\tau_f = c + \sigma\tan\phi$. Failure occurs when the Mohr circle of stresses touches this straight-line envelope.

Strength envelopes

1. Cohesive-frictional ($c$-$\phi$) soil (silty or sandy clay): the line has an intercept $c$ on the $\tau$-axis and slope $\phi$.
2. Cohesionless soil ($c=0$, e.g. clean sand): the line passes through the origin, $\tau_f = \sigma\tan\phi$.
3. Purely cohesive soil ($\phi=0$, e.g. saturated clay in undrained loading): the line is horizontal, $\tau_f = c$.

 tau            tau             tau
  |      /       |      /        |
  |    /         |    /          |---------- tau=c
  |  /  phi      |  /  phi       |
c |/             |/              |
  +---------s    +---------s     +---------s
 c-phi soil      sand (c=0)      pure clay
 tau=c+s*tan(phi) tau=s*tan(phi) (phi=0)

Limitations of the Mohr-Coulomb theory:

1. Intermediate principal stress is ignored. The criterion depends only on $\sigma_1$ and $\sigma_3$, but tests show $\sigma_2$ has some influence.
2. Straight-line envelope is an approximation. The true envelope is usually curved, especially at very low or very high stresses, so $c$ and $\phi$ are valid only for the stress range tested.
3. $c$ and $\phi$ are not fundamental soil constants. They depend on drainage condition, rate of loading, stress path, density, water content, and the test type.
4. Does not model stress-strain behaviour. It gives only the strength at failure, not the strains required to reach it.
5. Strain softening and residual strength are not included. Peak strength may not be available along an existing slip surface.
6. Anisotropy, fissures and structure are neglected; the soil is assumed isotropic and homogeneous.
7. Time effects such as creep are not considered.
8. Tension is not covered properly. The envelope extended to negative normal stress overestimates tensile strength.

Major and minor principal stresses

At any point in a stressed soil there are three mutually perpendicular planes on which the shear stress is zero. These are the principal planes, and the normal stresses on them are the principal stresses.

• Major principal stress $\sigma_1$: the largest of the three normal stresses.
• Minor principal stress $\sigma_3$: the smallest of the three normal stresses.
• The third is the intermediate principal stress $\sigma_2$.

Effect of increasing $\sigma_1$ with $\sigma_3$ constant

The deviator stress $\sigma_1 - \sigma_3$ increases, so the Mohr circle grows larger (its centre moves right and its radius increases). The maximum shear stress, $\frac{\sigma_1-\sigma_3}{2}$, also increases. When the circle becomes tangent to the Mohr-Coulomb envelope, the soil fails. If $\sigma_1$ is raised further, the soil would be past failure, which is impossible in a real soil: the stress state cannot go beyond the envelope.

Mohr circle at failure for a soil with only $\phi$ (cohesionless)

The failure line passes through the origin, and the circle is tangent to it:

 tau       /  tau = sigma tan(phi)
  |      T/
  |     /.-''-.
  |    /'      '.
  |  / phi  C    \
  +--+------+-----+---- sigma
     0     s3    s1

From the right-angled triangle $OTC$:

$$\sin\phi = \frac{\sigma_1-\sigma_3}{\sigma_1+\sigma_3}$$

and the failure plane is at $45^\circ+\phi/2$ to the major principal plane.

Laboratory shear strength tests: direct shear test, triaxial test (UU, CU, CD), unconfined compression test, and laboratory vane shear test.

Shear strength in the direct shear test

A soil specimen in a split box (commonly $60\times60$ mm) is loaded with a constant normal load $N$. The upper half is pushed horizontally until failure along the horizontal plane between the two halves. The horizontal force at failure is $T_f$ (read from the proving ring).

$$\sigma = \frac{N}{A}, \qquad \tau_f = \frac{T_f}{A}$$

where $A$ is the area of the specimen (corrected for the reduced contact area as the box moves, if required).

The test is repeated at three or more different normal stresses. The points $(\sigma, \tau_f)$ are plotted and a best-fit straight line is drawn. Its intercept on the $\tau$-axis is the cohesion $c$ and its slope is $\phi$:

$$\tau_f = c + \sigma\tan\phi$$

For sand, $c = 0$ and $\phi = \tan^{-1}(\tau_f/\sigma)$.

Drained shear strength is the strength mobilised when the soil is sheared slowly enough that water can drain freely, so no excess pore pressure develops. It is expressed in effective stress. Undrained shear strength is the strength when the soil is sheared quickly with no drainage, so volume cannot change and excess pore pressure develops.

The triaxial test has these advantages over the direct shear test:

1. Drainage can be controlled. UU, CU and CD tests are possible, and pore pressure can be measured. In direct shear, drainage cannot be controlled, so effective stress parameters are hard to obtain.
2. Failure plane is not forced. The sample fails along its weakest plane, while in direct shear the plane is fixed horizontally.
3. Uniform stress and strain in the specimen, so results are more reliable. Direct shear has stress concentration at the edges and progressive failure.
4. Principal stresses are known exactly. A complete Mohr circle can be drawn from $\sigma_1$ and $\sigma_3$.
5. Stress path can be followed; different stress paths such as compression or extension can be applied.
6. Volume change can be measured during shear, as well as strains.
7. Large specimen sizes can be used for coarse soil, and the test suits all soil types.
8. Stresses can be varied independently, which allows tests under in-situ stress conditions.

The triaxial compression test applies an all-round cell pressure $\sigma_3$ to a cylindrical specimen and then increases the axial stress until the sample fails.

Advantages

• Drainage conditions can be controlled and pore pressure measured.
• Failure takes place on the weakest plane; stresses are fairly uniform.
• Suitable for all soil types and for different stress paths.
• Effective and total stress parameters can both be obtained.
• Volume change and strain can be recorded.

Disadvantages

• Equipment is costly, and the test needs a skilled operator.
• CD and CU tests take a long time (days for clays).
• Stress state is axisymmetric ($\sigma_2 = \sigma_3$), which is not the real condition for plane-strain problems such as strip footings and slopes.
• Sample disturbance during preparation, and end restraint from the platens.
• Preparation of an undisturbed sample is difficult in sand.

Procedure

1. Prepare a cylindrical specimen (typically 38 mm diameter, 76 mm long), cover it with a rubber membrane and place it between porous discs on the pedestal.
2. Fill the cell with water and apply the cell pressure $\sigma_3$.
3. For CU and CD, allow consolidation under $\sigma_3$ with drainage open; for UU, keep drainage closed.
4. Apply axial load at a constant rate of strain until failure (or 15 to 20% strain), recording load, deformation and, if required, pore pressure $u$.
5. Repeat on at least three identical specimens at different cell pressures.

Computing the shear parameters

For each test, the deviator stress at failure is

$$\sigma_d = \sigma_1 - \sigma_3 = \frac{P}{A_c}, \qquad A_c = \frac{A_0}{1-\varepsilon}$$

where $P$ is the axial load and $A_c$ is the corrected area.

1. Calculate $\sigma_1 = \sigma_3 + \sigma_d$ for each test (use $\sigma' = \sigma - u$ for effective stress).
2. Draw a Mohr circle for each test on the $\sigma$-$\tau$ plane with diameter from $\sigma_3$ to $\sigma_1$.
3. Draw the common tangent to the circles, which is the failure envelope.
4. Read the intercept on the $\tau$-axis as $c$ and the slope as $\phi$ (or $c'$, $\phi'$ for effective stress).

Alternatively, $\sin\phi = \dfrac{\sigma_1 - \sigma_3}{\sigma_1+\sigma_3+2c\cot\phi}$, and for $c=0$, $\sin\phi = \dfrac{\sigma_1-\sigma_3}{\sigma_1+\sigma_3}$.

The three triaxial tests represent three field situations, depending on how fast the load is applied compared with the rate at which water can drain from the soil.

Key idea: clays under quick loading behave undrained (UU), clays under slow loading behave drained (CD), and sands drain quickly so they are almost always treated as drained.

Triaxial tests by drainage condition

1. Unconsolidated undrained (UU) test: no drainage during consolidation or shearing.
2. Consolidated undrained (CU) test: drainage during consolidation, none during shearing (pore pressure may be measured).
3. Consolidated drained (CD) test: drainage allowed during both consolidation and shearing.

Unconfined compressive strength vs undrained shear strength

UU test (unconsolidated undrained)

• Gives $c_u$ with $\phi_u \approx 0$ for saturated clay.
• Used where loading is fast and the clay has no time to drain: stability of foundations and embankments on soft clay at the end of construction, bearing capacity in the short term, and temporary excavations in clay.

CU test (consolidated undrained)

• Gives $c_{cu}$ and $\phi_{cu}$ (total stress) and, with pore pressure measurement, $c'$ and $\phi'$ (effective stress).
• Used where soil is first consolidated and then loaded quickly: rapid drawdown of an earth dam after steady seepage, stage-constructed embankments, and sudden loading of already-consolidated clay under a structure.

CD test (consolidated drained)

• Gives $c_d$ and $\phi_d$ (about equal to $c'$, $\phi'$).
• Used for long-term stability when excess pore pressure has dissipated: natural slopes and cuttings, retaining walls and drained foundations, and for sand and gravel, which drain quickly.

Soil type

The unconfined compression test is conducted on saturated, cohesive soils (clays) that can stand without lateral support, using undisturbed samples. It is not suitable for dry sands, gravels, or fissured clays, which cannot hold their shape without confining pressure.

Determination of strength parameters using Mohr circle

In the test, the lateral pressure is zero ($\sigma_3 = 0$) and the axial stress is increased to failure:

$$\sigma_1 = q_u = \frac{P}{A_c}, \qquad A_c = \frac{A_0}{1-\varepsilon}$$

The Mohr circle at failure passes through the origin with diameter $q_u$. For saturated clay, the undrained envelope is horizontal ($\phi_u = 0$), so the circle touches it at the top:

 tau
  |        ..--..
c_u ------*'     '*-------- envelope (phi_u = 0)
  |      /   C    \
  +-----+----+-----+----- sigma
  0 (s3=0)         q_u = s1

$$c_u = \frac{q_u}{2}$$

The failure plane would be at $45^\circ + \phi_u/2 = 45^\circ$ to the horizontal. If the sample is partly saturated or $\phi_u \neq 0$ is expected, one more test with a different confining pressure is required to find both $c_u$ and $\phi_u$. The test is quick: it gives only the undrained cohesion $c_u = q_u/2$.

Unconfined compression test (UCS)

Plot axial stress $\sigma_1$ against axial strain; the peak is $q_u$. The Mohr circle drawn on the $\sigma$-$\tau$ plane passes through the origin ($\sigma_3 = 0$) with $\sigma_1 = q_u$, and the envelope is a horizontal line tangent to the circle for $\phi_u = 0$:

 UCS stress-strain          Mohr circle
 s1                         tau
  |   ___ qu                 |   ..--..
  |  /   \                 cu |--*'    '*--
  | /                         | /   C    \
  +--------- strain          -+-+--------+--- sigma
                              0 (s3=0)    qu

Direct shear test

Plot shear stress against horizontal displacement for each normal stress to get the peak $\tau_f$. Then plot $\tau_f$ against normal stress $\sigma$ for three or more tests and draw the best-fit straight line:

 tau_f            /
   |            /  <- slope = phi
   |          /
 c |--------/
   |      .
   +--------------- sigma
   s1  s2  s3

Intercept $= c$; slope angle $= \phi$. For the same clay, the UCS gives $c_u = q_u/2$ (total stress, quick), while the direct shear (slow) gives $c$ and $\phi$ under drained conditions.

In a UU triaxial test, a specimen is sheared without drainage and without prior consolidation under a cell pressure $\sigma_3$. The unconfined compression test is the same test with cell pressure equal to zero ($\sigma_3 = 0$).

For saturated clay, the undrained strength is independent of the confining pressure ($\phi_u = 0$), so a sample tested without any cell pressure gives the same Mohr circle diameter as one tested with a cell pressure, shifted along the $\sigma$-axis:

$$q_u = \sigma_1 - 0 = \sigma_1 - \sigma_3 = 2c_u$$

Both tests also shear the sample quickly with no drainage (no change in water content). Hence UCS is a special case of the UU test with $\sigma_3 = 0$.

In a drained direct shear test on sand under a constant normal stress:

• Dense sand shows a distinct peak shear stress at small displacement, then drops to a lower (residual or ultimate) value. It first contracts slightly and then dilates (expands), so the height of the specimen increases.
• Loose sand shows shear stress rising gradually to a maximum without a peak, and then staying nearly constant. It contracts continuously, so the specimen height decreases.

 Shear stress
 tau
  |      ,-.
  |    ,'   `--.__  dense (peak, then drops)
  |   /     .-----------  loose (gradual rise)
  |  / ..-''
  | /.'
  +-------------------- shear displacement
              Both reach about the same
              ultimate value (same phi_cv)

 Change in height (expansion up)
  dH
  |            ___ dense (dilates)
  |         ,-'
  |  ___.--'
  +----`.__----------- shear displacement
  |       `-.__ loose (contracts)
  |

The peak friction angle of dense sand is greater than that of loose sand, but both reach the same ultimate (critical state) strength at large strains.

A stress path is the locus of successive stress states of a soil element during loading, plotted as a line on a stress diagram. It shows how stress changes from the initial state to failure.

It is usually plotted in $p$-$q$ space, where $p = \frac{\sigma_1+\sigma_3}{2}$ and $q = \frac{\sigma_1-\sigma_3}{2}$, with the $K_f$ line (failure line) $q = p\sin\phi + c\cos\phi$. Each point represents a Mohr circle, so stress paths avoid drawing many circles. Total and effective stress paths differ by the pore pressure.

Limitations of the direct shear test:

1. Drainage cannot be controlled. Pore pressure cannot be measured, so the test cannot give true undrained or effective stress parameters for clays.
2. Failure plane is fixed (horizontal). The soil may not fail along its weakest plane.
3. Non-uniform stress and strain distribution over the failure plane, with stress concentration at the edges and progressive failure.
4. Area of contact decreases as the upper box moves, so the shear stress is not exactly $T/A$ and a correction is needed.
5. Principal planes rotate during shear and the major principal stress direction is not known; a full Mohr circle cannot be drawn.
6. Small specimen size limits use for coarse-grained soil and gives a small, non-representative sample.
7. Side friction between the box and the soil, and the effect of the rigid boundaries, introduce errors.
8. Difficult to apply other stress paths, and the volume change is hard to measure accurately.

Step 1: Principal stresses at failure ($\sigma_1 = \sigma_3 + \sigma_d$)

Step 2: Mohr circles and envelope. Draw each circle with its centre on the $\sigma$-axis and radius as in the table, then draw the common tangent.

 tau
  |            ___..--''''-.
  |        ,-''   ___...--'' `.
  |     ,-'  _.--'' .--''  .   \
  |   ,'  .-'  __,-'  _.-'  \   \
  |  / .'.-' ,-'  _.-'   phi  |  |
  +--+---------------------------+-- sigma
    0   300  400  500   1175 1560 1960

The tangent passes (almost) through the origin, so $c \approx 0$ (a cohesionless soil).

Step 3: Angle of internal friction

$$\sin\phi = \frac{\sigma_1-\sigma_3}{\sigma_1+\sigma_3}$$ $$\begin{aligned} \text{Test 1: } \sin\phi &= \frac{875}{1475} = 0.593 \;\Rightarrow\; \phi = 36.4^\circ \\ \text{Test 2: } \sin\phi &= \frac{1160}{1960} = 0.592 \;\Rightarrow\; \phi = 36.3^\circ \\ \text{Test 3: } \sin\phi &= \frac{1460}{2460} = 0.594 \;\Rightarrow\; \phi = 36.4^\circ \end{aligned}$$

Answer: $c \approx 0$ and $\phi \approx 36.4^\circ$ (about $36^\circ$).

For normally consolidated clay, $c' = 0$ and $c_{cu} = 0$, so $\sin\phi = \dfrac{\sigma_1-\sigma_3}{\sigma_1+\sigma_3}$.

(i) $\phi'$ from the drained test

$\sigma_3' = 300$, $\sigma_1' = 300+650 = 950$ kPa.

$$\sin\phi' = \frac{650}{300+950} = \frac{650}{1250} = 0.52 \;\Rightarrow\; \phi' = 31.3^\circ$$

(ii) $\phi$ from the CU test (total stress)

$\sigma_3 = 200$, $\sigma_1 = 200+250 = 450$ kPa.

$$\sin\phi_{cu} = \frac{250}{200+450} = \frac{250}{650} = 0.385 \;\Rightarrow\; \phi_{cu} = 22.6^\circ$$

(iii) Pore pressure at failure in the CU test

Assume the same effective friction angle holds, $\phi' = 31.3^\circ$, and $c'=0$. At failure the effective Mohr circle has the same deviator stress, 250 kPa:

$$\sin\phi' = \frac{\sigma_1'-\sigma_3'}{\sigma_1'+\sigma_3'} = \frac{250}{2\sigma_3' + 250} = 0.52$$ $$2\sigma_3' + 250 = \frac{250}{0.52} = 480.8 \;\Rightarrow\; \sigma_3' = 115.4\ \text{kPa}$$ $$u_f = \sigma_3 - \sigma_3' = 200 - 115.4 = 84.6\ \text{kPa}$$

Answer: (i) $\phi' = 31.3^\circ$; (ii) $\phi_{cu} = 22.6^\circ$; (iii) $u_f \approx 84.6$ kPa.

Step 1: Effective principal stresses at failure

$\sigma_3' = \sigma_3 - u$ and $\sigma_1' = \sigma_3' + \sigma_d$.

Step 2: Envelope. Draw the four effective-stress circles and the common tangent. Alternatively, fit a straight line through the points $(p', q)$ (the $K_f$ line), $q = a + p'\tan\alpha$. A least-squares fit gives:

$$\tan\alpha = 0.4421, \qquad a = 18.7\ \text{kPa}$$

Step 3: Parameters

$$\sin\phi' = \tan\alpha = 0.4421 \;\Rightarrow\; \phi' = 26.2^\circ$$ $$c' = \frac{a}{\cos\phi'} = \frac{18.7}{0.8975} = 20.8\ \text{kPa}$$

Answer: $c' \approx 21$ kN/m² and $\phi' \approx 26^\circ$ (in terms of effective stress).

Assumptions: $\gamma_w = 9.81$ kN/m³; the soil above the water table is partly saturated ($S_r = 0.5$); $c=0$ and capillary effects are neglected. Strength $= \sigma'\tan\phi$ on the horizontal plane.

Case 1: Water table at 3.5 m (below the 3 m depth, so there is no pore pressure at the plane)

$$\gamma = \frac{(G_s + S_r e)\gamma_w}{1+e} = \frac{(2.70 + 0.5\times0.50)(9.81)}{1.50} = 19.29\ \text{kN/m}^3$$ $$\sigma' = \sigma = 19.29\times 3 = 57.9\ \text{kPa}$$ $$s = \sigma'\tan\phi = 57.9\tan30^\circ = 33.4\ \text{kPa}$$

Case 2: Water table at the ground surface (fully saturated, $S_r = 1$)

$$\gamma_{sat} = \frac{(G_s + e)\gamma_w}{1+e} = \frac{(2.70+0.50)(9.81)}{1.50} = 20.93\ \text{kN/m}^3$$ $$\sigma' = (20.93 - 9.81)\times 3 = 33.4\ \text{kPa}$$ $$s = 33.4\tan30^\circ = 19.3\ \text{kPa}$$

Answer: shear strength is about 33.4 kPa with the water table at 3.5 m, and falls to about 19.3 kPa (a reduction of about 42%) when the water table rises to the ground surface.

Step 1: Principal stresses. $\sigma_1 = \sigma_3 + \sigma_d$, effective: $\sigma' = \sigma - u$.

Step 2: Plot. Draw three total stress circles (diameters 200 to 444, 300 to 614, 400 to 784) and three effective stress circles (145 to 389, 193 to 507, 241 to 625). Draw a common tangent for each set.

 tau
  |       _.-'''-._   total-stress envelope
  |   _.-'  _.-''-. `-.     (phi_cu, c_cu)
  |  /   .-'  effective  \   envelope is steeper
  +-+--+----+-----+----+---+---- sigma
   145 200  241   389  444 ...

Step 3: Parameters. Both sets of points fit straight lines of the form $\sigma_1 = N\sigma_3 + 2c\sqrt{N}$.

Total stress: slope $N = \dfrac{614-444}{300-200} = 1.70$, intercept $= 104$ kPa.

$$\phi_{cu} = \sin^{-1}\frac{N-1}{N+1} = \sin^{-1}\frac{0.70}{2.70} = 15.0^\circ, \qquad c_{cu} = \frac{104}{2\sqrt{1.70}} = 39.9\ \text{kPa}$$

Effective stress: slope $N' = \dfrac{507-389}{193-145} = 2.458$, intercept $= 32.5$ kPa.

$$\phi' = \sin^{-1}\frac{1.458}{3.458} = 24.9^\circ, \qquad c' = \frac{32.5}{2\sqrt{2.458}} = 10.4\ \text{kPa}$$

Answer: total stress $c_{cu} \approx 40$ kN/m², $\phi_{cu} \approx 15^\circ$; effective stress $c' \approx 10$ kN/m², $\phi' \approx 25^\circ$. (For an ideal normally consolidated clay the intercepts would be close to zero; the small cohesion shown here comes from the given data.)

Assumptions: $\gamma_w = 9.81$ kN/m³; sand above the water table is dry ($\gamma = \gamma_d = 17$); $c = 0$.

Void ratio

$$\gamma_d = \frac{G_s\gamma_w}{1+e} \;\Rightarrow\; e = \frac{2.7\times9.81}{17} - 1 = 0.558$$ $$\gamma_{sat} = \frac{(G_s+e)\gamma_w}{1+e} = \frac{(2.7+0.558)(9.81)}{1.558} = 20.51\ \text{kN/m}^3$$

Case 1: Water table at 2.5 m

$$\begin{aligned} \sigma' &= 17(2.5) + (20.51 - 9.81)(1.5) \\ &= 42.5 + 16.05 = 58.6\ \text{kPa} \end{aligned}$$ $$s = \sigma'\tan\phi = 58.6\tan35^\circ = 41.0\ \text{kPa}$$

Case 2: Water table at the surface

$$\sigma' = (20.51-9.81)(4) = 42.8\ \text{kPa}, \qquad s = 42.8\tan35^\circ = 30.0\ \text{kPa}$$

Change in shear strength: $30.0 - 41.0 = -11.0$ kPa.

Answer: shear strength is 41.0 kPa with the water table at 2.5 m; it falls by about 11.0 kPa (27%) to 30.0 kPa when the water table rises to the surface.

Data: $\sigma_3 = 200$, $\sigma_d = 350$, so $\sigma_1 = 550$ kN/m²; $u = 80$ kN/m²; $c = c' = 0$.

(i) Mohr circles

• Total stress: $\sigma_3 = 200$, $\sigma_1 = 550$; centre 375, radius 175.
• Effective stress: $\sigma_3' = 200-80 = 120$, $\sigma_1' = 550-80 = 470$; centre 295, radius 175.

The effective circle has the same size but is shifted left by $u = 80$. Draw both, with a line through the origin tangent to each:

 tau
  |       /  /
  |     /  /   . total circle
  |   /  / .-'''-.
  |  / /.'  .-'''-\.
  | //  eff.'      '
  +-+----+--+-------+-- sigma
   0  120 200  470  550

(ii) Friction angles

$$\sin\phi = \frac{\sigma_1-\sigma_3}{\sigma_1+\sigma_3} = \frac{350}{750} = 0.467 \;\Rightarrow\; \phi = 27.8^\circ$$ $$\sin\phi' = \frac{\sigma_1'-\sigma_3'}{\sigma_1'+\sigma_3'} = \frac{350}{590} = 0.593 \;\Rightarrow\; \phi' = 36.4^\circ$$

(iii) Direction of the failure plane

The failure plane is inclined at $\theta = 45^\circ + \phi'/2$ to the major principal plane (the horizontal plane here, since the vertical stress is the major principal stress):

$$\theta = 45^\circ + \frac{36.4^\circ}{2} = 63.2^\circ \text{ to the horizontal}$$

It makes $26.8^\circ$ with the vertical axis of the specimen. (Using the total stress angle, $45^\circ + 27.8^\circ/2 = 58.9^\circ$.)

Answer: $\phi = 27.8^\circ$, $\phi' = 36.4^\circ$; failure plane at about $63.2^\circ$ to the horizontal.

For dry cohesionless soil, $c = 0$ and

$$\sigma_1 = \sigma_3\tan^2\left(45^\circ + \frac{\phi}{2}\right)$$

Given: $\phi = 36^\circ$, $\sigma_3 = 100$ kN/m².

$$\tan^2(63^\circ) = (1.9626)^2 = 3.852$$ $$\sigma_1 = 100 \times 3.852 = 385.2\ \text{kN/m}^2$$ $$\sigma_d = \sigma_1 - \sigma_3 = 385.2 - 100 = 285.2\ \text{kN/m}^2$$

Check: $\sin\phi = \frac{285.2}{385.2+100} = 0.588 = \sin36^\circ$.

Answer: deviator stress at failure $\approx 285\ \text{kN/m}^2$.

For a cohesionless soil ($c=0$), the failure circle touches a line through the origin.

Test 1: $\sigma_3 = 100$ kPa, $\sigma_d = 200$ kPa, so $\sigma_1 = 300$ kPa.

$$\sin\phi = \frac{\sigma_1-\sigma_3}{\sigma_1+\sigma_3} = \frac{200}{400} = 0.5 \;\Rightarrow\; \phi = 30^\circ$$

Test 2: $\sigma_3 = 200$ kPa.

$$\sigma_1 = \sigma_3\tan^2\left(45^\circ+\frac{30^\circ}{2}\right) = 200 \times 3 = 600\ \text{kPa}$$ $$\sigma_d = 600 - 200 = 400\ \text{kPa}$$

(The deviator stress is proportional to $\sigma_3$ for cohesionless soil, so doubling $\sigma_3$ doubles $\sigma_d$.)

Mohr circles and envelope

 tau        / tau = sigma tan30
  |       /
  |     /   ..-'''-.      circle 2: 200 to 600
  |   /  .-'.-'''-.  \
  | /  .'  /  c1   \  |   circle 1: 100 to 300
  +-+--+----+-+-----+---- sigma
   0  100  200 300  600

Answer: $\phi = 30^\circ$; deviator stress at failure for $\sigma_3 = 200$ kPa is 400 kPa.

Given: dry sand ($c=0$), deviator stress at failure $\sigma_1-\sigma_3 = 500$ kN/m². The failure plane makes $25^\circ$ with the axis of the sample.

Step 1: Angle of the failure plane. The axis of the sample is the direction of the major principal stress $\sigma_1$, so the failure plane makes $25^\circ$ with the direction of $\sigma_1$ and hence $90^\circ - 25^\circ = 65^\circ$ with the major principal plane (the horizontal plane). The theoretical failure angle is

$$\theta = 45^\circ + \frac{\phi}{2} = 65^\circ \;\Rightarrow\; \phi = 40^\circ$$

Step 2: Flow value

$$N_\phi = \tan^2\left(45^\circ+\frac{\phi}{2}\right) = \tan^2 65^\circ = 4.599$$

Step 3: Lateral pressure. For $c=0$:

$$\sigma_1 = N_\phi\sigma_3 \;\Rightarrow\; \sigma_1 - \sigma_3 = \sigma_3(N_\phi - 1)$$ $$\sigma_3 = \frac{500}{4.599 - 1} = \frac{500}{3.599} = 138.9\ \text{kN/m}^2$$

Check: $\sigma_1 = 638.9$, $\sin\phi = \frac{500}{638.9+138.9} = 0.643 = \sin40^\circ$.

Answer: $\phi = 40^\circ$ and the lateral pressure $\sigma_3 \approx 139\ \text{kN/m}^2$.

Data: $\sigma_3 = 200$, $\sigma_1 = 550$, $u = 80$ kN/m². Deviator stress $= 350$ kN/m². Effective stresses: $\sigma_3' = 200-80 = 120$, $\sigma_1' = 550-80 = 470$ kN/m².

i) Mohr circle, total stress

Centre $= \frac{200+550}{2} = 375$, radius $= \frac{550-200}{2} = 175$. The circle cuts the $\sigma$-axis at 200 and 550.

ii) Mohr circle, effective stress

Centre $= \frac{120+470}{2} = 295$, radius $= 175$. The circle cuts the $\sigma$-axis at 120 and 470. It is the total stress circle shifted left by $u = 80$.

 tau
  |          /     /
  |        /     /   (effective circle)
  |      /     /  .-'''-.   (total circle)
  |    /     / .-'.-'''-. \
  |  /     /.-'  '       ' |
  +-+--+---+----+-------+-+-- sigma
   0  120 200  470     550

iii) Effective friction angle

For $c' = 0$:

$$\sin\phi' = \frac{\sigma_1'-\sigma_3'}{\sigma_1'+\sigma_3'} = \frac{350}{590} = 0.593 \;\Rightarrow\; \phi' = 36.4^\circ$$

(For information, the total stress angle is $\sin\phi = 350/750$, $\phi = 27.8^\circ$.)

iv) Direction of the failure plane

The failure plane is inclined to the major principal plane (horizontal) at

$$\theta = 45^\circ + \frac{\phi'}{2} = 45^\circ + 18.2^\circ = 63.2^\circ$$

So the plane is at about $63.2^\circ$ to the horizontal, which is $26.8^\circ$ to the axis of the specimen.

Answer: $\phi' \approx 36.4^\circ$; failure plane at about $63^\circ$ to the horizontal.