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Direct Shear Test

In a direct shear test, a soil specimen is placed in a split shear box. A normal stress is applied first, then one half of the box is moved relative to the other until the specimen fails in shear.

$$\sigma = \frac{P}{A}$$
$$\tau = \frac{V}{A}$$

$P$ is normal load, $V$ is shear force at failure, and $A$ is cross-sectional area of the shear plane.

Mohr-Coulomb Shear Strength

The shear strength of soil is commonly represented by the Mohr-Coulomb failure equation.

$$\tau_f = c + \sigma\tan\phi$$
$$\tau_f = \sigma\tan\phi \quad \text{for dry clean sand}$$
$$\phi = \tan^{-1}\left(\frac{\tau_f-c}{\sigma}\right)$$

Here $c$ is cohesion and $\phi$ is the angle of internal friction.

Triaxial Tests

In a triaxial test, the specimen is enclosed in a membrane, subjected to confining pressure, then loaded axially until failure. The axial stress increment at failure is called deviator stress.

$$\sigma_1 = \sigma_3 + \Delta\sigma_d$$
$$\sin\phi = \frac{\sigma_1-\sigma_3}{\sigma_1+\sigma_3} \quad \text{for } c=0$$
$$\theta = 45^\circ + \frac{\phi}{2}$$

CD tests allow drainage during consolidation and shearing. CU tests consolidate first, then shear without drainage. UU tests do not allow drainage during either stage.

Undrained and Unconfined Compression Strength

For saturated clay in UU loading, the total-stress friction angle is often taken as zero. The undrained shear strength is the radius of Mohr's circle.

$$\phi_u = 0$$
$$C_u = \frac{\sigma_1-\sigma_3}{2}$$
$$q_u = \sigma_1 \quad \text{when } \sigma_3=0$$
$$C_u = \frac{q_u}{2}$$

Problem: Direct Shear on Dry Sand

A dry sand sample is tested in direct shear. The shear box is circular with diameter 50 mm. Normal load is 200 N and failure shear force is 130 N. Determine $\phi$.

$$A = \frac{\pi(50)^2}{4}=1963.5 \text{ mm}^2$$
$$\tan\phi = \frac{\tau}{\sigma} = \frac{V/A}{P/A} = \frac{130}{200}$$
$$\phi = \tan^{-1}(0.65)=33.02^\circ$$

Answer: $\phi = 33.02^\circ$.

Problem: Direct Shear Table Workflow

A direct shear test gives several combinations of normal force $P$ and shear force $V$ at failure. Compute cohesion, friction angle, and $V$ for a new value of $P$.

The numerical table is embedded as an image in the PDF. Use this process once the table values are read:

$$\sigma_i = \frac{P_i}{A}$$
$$\tau_i = \frac{V_i}{A}$$
$$\tau = c+\sigma\tan\phi$$

Plot $\tau$ versus $\sigma$ or compute the slope from two points. The intercept is $c$ and the slope is $\tan\phi$.

Problem: Rectangular Direct Shear Sample

A dry sand sample 60 mm by 60 mm by 25 mm high is tested in direct shear. Normal stress is 100 kPa and shear force at failure is 300 N. Determine the angle of friction.

$$A = 0.06(0.06)=0.0036 \text{ m}^2$$
$$\tau = \frac{300}{0.0036}=83.33 \text{ kPa}$$
$$\phi=\tan^{-1}\left(\frac{83.33}{100}\right)=39.81^\circ$$

Answer: $\phi=39.81^\circ$.

Problem: Undrained Shear Box Workflow

A series of undrained shear box tests gives normal loads and shear loads at failure. Determine cohesion of the soil sample.

The load table is embedded as an image in the PDF. Convert each load to stress using the given box area, then plot the failure envelope.

$$\sigma = \frac{P}{A}$$
$$\tau = \frac{V}{A}$$
$$c = \tau \text{ at } \sigma=0$$

Problem: Dry Sand Shear Stress

A dry sand sample has normal stress 200 kPa and fails at shear stress 135 kPa. Determine $\phi$, the failure shear stress for normal stress 145 kPa, and shear stress at 5 m depth if $\gamma=15.8$ kN/m3.

$$\phi=\tan^{-1}\left(\frac{135}{200}\right)=34.02^\circ$$
$$\tau_f = 145\tan34.02^\circ=97.88 \text{ kPa}$$
$$\sigma_v = 15.8(5)=79 \text{ kPa}$$
$$\tau_f = 79\tan34.02^\circ=53.33 \text{ kPa}$$

Answer: $\phi=34.02^\circ$, $\tau=97.88$ kPa, and $\tau=53.33$ kPa at 5 m depth.

Problem: Triaxial Stress on Failure Plane

A drained sand sample has normal stress 50 kPa and shear stress 30 kPa on the failure plane. Determine internal friction angle, failure plane angle with respect to the horizontal plane, and axial stress applied.

$$\phi=\tan^{-1}\left(\frac{30}{50}\right)=30.96^\circ$$
$$\theta=45^\circ+\frac{30.96^\circ}{2}=60.48^\circ$$
$$C = 50+30\tan30.96^\circ=68.0 \text{ kPa}$$
$$R=\sqrt{(68-50)^2+30^2}=34.99 \text{ kPa}$$
$$\Delta\sigma_d=2R=69.97 \text{ kPa}$$

Answer: $\phi=30.96^\circ$, $\theta=60.48^\circ$, and axial/deviator stress is about 69.97 kPa.

Problem: Cohesive Soil Cell Pressure

A cohesive soil has $\phi=28^\circ$ and $c=31$ kPa. If maximum shearing stress is 65 kPa, compute lateral pressure in the cell at failure.

$$R = 65 \text{ kPa}$$
$$R = c\cos\phi + C\sin\phi$$
$$65 = 31\cos28^\circ + C\sin28^\circ$$
$$C = 80.15 \text{ kPa}$$
$$\sigma_3 = C-R = 80.15-65=15.15 \text{ kPa}$$

Answer: Lateral cell pressure is 15.15 kPa.

Problem: Maximum Principal Stress

In a triaxial test for normally consolidated soil, normal stress at failure is 475 kPa and shear stress at failure is 350 kPa. Compute the maximum principal stress at failure.

$$\phi=\tan^{-1}\left(\frac{350}{475}\right)=36.38^\circ$$
$$C = 475+350\tan36.38^\circ=732.89 \text{ kPa}$$
$$R=\sqrt{(732.89-475)^2+350^2}=434.76 \text{ kPa}$$
$$\sigma_1=C+R=1167.65 \text{ kPa}$$

Answer: $\sigma_1=1167.65$ kPa.

Problem: Normal Stress at Maximum Shear

A cohesive soil has $\phi=28^\circ$ and $c=30$ kPa. If maximum shearing stress is 70 kPa, compute the normal stress at the point of maximum shear.

$$R=70 \text{ kPa}$$
$$70=30\cos28^\circ+C\sin28^\circ$$
$$C=92.68 \text{ kPa}$$

The normal stress at maximum shear is the center of Mohr's circle.

Answer: 92.68 kPa.

Problem: Cohesionless Triaxial Test

In a triaxial test on cohesionless soil, chamber pressure is 10 kPa and failure occurs when major compressive stress reaches 30 kPa. Compute the angle of internal friction.

$$\sin\phi=\frac{\sigma_1-\sigma_3}{\sigma_1+\sigma_3}$$
$$\sin\phi=\frac{30-10}{30+10}=0.50$$
$$\phi=30^\circ$$

Answer: $\phi=30^\circ$.

Problem: Unconfined Compression Strength

During an unconsolidated undrained test, the principal stress at failure is 114 kPa. Compute undrained cohesion for the unconfined compression test.

$$C_u=\frac{q_u}{2}$$
$$C_u=\frac{114}{2}=57 \text{ kPa}$$

Answer: 57 kPa.

Problem: Clay Cohesion from Failure Load

An unconfined compression test was conducted on a soil sample 50 mm in diameter. Failure load was 66 N. Find cohesion strength of clay.

$$A=\frac{\pi(0.05)^2}{4}=0.0019635 \text{ m}^2$$
$$q_u=\frac{66}{0.0019635}=33.61 \text{ kPa}$$
$$C_u=\frac{33.61}{2}=16.81 \text{ kPa}$$

Answer: 16.81 kPa.

Problem: Pore Water Pressure in CU Test

For fully saturated clay, $c'=15$ kPa and $\phi'=29^\circ$. In a CU triaxial test, all-around pressure is 100 kPa and principal stress difference at failure is 170 kPa. Find expected pore water pressure at failure.

$$\sigma_3'=100-u$$
$$\sigma_1'=270-u$$
$$R=\frac{\sigma_1'-\sigma_3'}{2}=85 \text{ kPa}$$
$$C=\frac{\sigma_1'+\sigma_3'}{2}=185-u$$
$$R=c'\cos\phi'+C\sin\phi'$$
$$85=15\cos29^\circ+(185-u)\sin29^\circ$$
$$u=36.73 \text{ kPa}$$

Answer: Expected pore water pressure is 36.73 kPa.

Problem: Drained Friction Angle from CU Table

Loose sand samples were tested under consolidated undrained conditions. Failure stresses and excess pore water pressures are tabulated in the PDF. Compute drained friction angle.

The numerical table is embedded as an image. Use the effective principal stresses at failure.

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

Problem: Shear Strength from SPT

Using the SPT correlation table for cohesive soils, estimate approximate shear strength if blow count $N=6$.

For $N=6$, the soil is in the medium range, with $q_u$ approximately between 47.8 kPa and 95.7 kPa.

$$q_u \approx \frac{47.8+95.7}{2}=71.75 \text{ kPa}$$
$$C_u \approx \frac{q_u}{2}=\frac{71.75}{2}=35.9 \text{ kPa}$$

Answer: Approximate shear strength is about 35.9 kPa.

Concept: Mohr-Coulomb Failure Criterion

The Mohr-Coulomb criterion defines the shear stress at failure on any plane in a soil mass:

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

where $c'$ is effective cohesion and $\phi'$ is effective friction angle. For clean sands $c'=0$; for saturated clays under undrained loading $\phi_u=0$ and $\tau_f=S_u$. In a triaxial test the failure plane is inclined at $\alpha_f=45+\phi/2$ from the major principal plane. The Mohr circle at failure gives:

$$\sin\phi = \frac{(\sigma_1-\sigma_3)/2}{(\sigma_1+\sigma_3)/2} = \frac{\sigma_1-\sigma_3}{\sigma_1+\sigma_3}$$

In a direct shear test the failure plane is forced horizontal, so normal and shear stresses are read directly. Plotting results from multiple tests gives the Mohr-Coulomb envelope: slope $=\tan\phi$, intercept $=c$. Sensitivity $S_t = S_{u,\text{undisturbed}}/S_{u,\text{remolded}}$ measures how much a clay loses strength when disturbed.

Problem: Direct Shear Test — Find $c$ and $\phi$

Three direct shear tests on a dry sand specimen yield the following failure data: Test 1 — normal stress 50 kPa, shear stress 29 kPa; Test 2 — normal stress 100 kPa, shear stress 58 kPa; Test 3 — normal stress 150 kPa, shear stress 87 kPa. Determine the cohesion $c$ and friction angle $\phi$.

Fit a straight line $\tau_f = c + \sigma_n\tan\phi$ through the three points. Use the slope from two end-points:

$$\tan\phi = \frac{87 - 29}{150 - 50} = \frac{58}{100} = 0.580$$
$$\phi = \arctan(0.580) = 30.1°$$

Intercept from Test 1: $29 = 0.580(50) + c \Rightarrow c = 29 - 29 = 0$ kPa.

$$c = 0\text{ kPa},\quad \phi = 30.1°$$

Answer: $c = 0$ kPa and $\phi = 30.1°$ (clean sand, friction only).

Problem: CU Triaxial — Failure Plane and Shear Stress

A consolidated-undrained triaxial test on saturated sand ($c=0$) gives: cell pressure $\sigma_3=100$ kPa, deviator stress at failure $\sigma_1-\sigma_3=280$ kPa. Find (a) friction angle $\phi$, (b) angle of failure plane from horizontal, and (c) shear stress on the failure plane.

$$\sigma_1 = 100 + 280 = 380\text{ kPa}$$
$$\sin\phi = \frac{\sigma_1-\sigma_3}{\sigma_1+\sigma_3} = \frac{280}{480} = 0.5833 \Rightarrow \phi = 35.7°$$
$$\alpha_f = 45 + \frac{\phi}{2} = 45 + 17.85 = 62.85°$$

Shear stress on the failure plane using Mohr circle (radius $R=140$ kPa):

$$\tau_f = R\sin(2\alpha_f) = 140\sin(125.7°) = 140(0.813) = 113.8\text{ kPa}$$

Answer: $\phi=35.7°$, failure plane at $62.85°$ from horizontal, $\tau_f=113.8$ kPa.

Problem: Three CD Triaxial Tests — Determine $c'$ and $\phi'$

Three drained triaxial tests on a sandy clay give: Test 1 — $\sigma_3=40$ kPa, $\sigma_1=200$ kPa; Test 2 — $\sigma_3=80$ kPa, $\sigma_1=320$ kPa; Test 3 — $\sigma_3=160$ kPa, $\sigma_1=560$ kPa. Find the effective strength parameters $c'$ and $\phi'$.

Express as $\sigma_1 = N_\phi\,\sigma_3 + K$ where $N_\phi=\tan^2(45+\phi/2)$ and $K=2c\sqrt{N_\phi}$. Subtract equations for Tests 1 and 2:

$$N_\phi = \frac{320 - 200}{80 - 40} = \frac{120}{40} = 3.0$$
$$\tan^2\!\left(45+\frac{\phi'}{2}\right) = 3.0 \Rightarrow 45+\frac{\phi'}{2} = 60° \Rightarrow \phi' = 30°$$

$K = 200 - 40(3.0) = 80$, and $K = 2c\tan(60°) = 2c(1.732)$:

$$c' = \frac{80}{2(1.732)} = 23.1\text{ kPa}$$

Verify with Test 3: $160(3)+80=560$ ✓

Answer: $c'=23.1$ kPa and $\phi'=30°$.

Problem: Field Vane Shear — Torque to Undrained Strength

A vane shear test uses a vane with height $H=130$ mm and diameter $D=65$ mm. Torque at failure is $T=38$ N·m. Calculate undrained shear strength $S_u$, assuming equal shear mobilization on the cylindrical surface and both end caps.

$$T = S_u\pi\!\left(\frac{D^2H}{2} + \frac{D^3}{6}\right)$$
$$\frac{D^2H}{2} = \frac{(0.065)^2(0.130)}{2} = 2.746\times10^{-4}\text{ m}^3$$
$$\frac{D^3}{6} = \frac{(0.065)^3}{6} = 4.577\times10^{-5}\text{ m}^3$$
$$S_u = \frac{38}{\pi(2.746+0.4577)\times10^{-4}} = \frac{38}{\pi(3.204\times10^{-4})} = \frac{38}{1.007\times10^{-3}}$$
$$S_u = 37{,}740\text{ Pa} \approx 37.7\text{ kPa}$$

Answer: $S_u \approx 37.7$ kPa.

Problem: Clay Sensitivity from Unconfined Compression

An undisturbed clay sample has unconfined compressive strength $q_u=144$ kPa. After remolding at the same water content, $q_{u,\text{rem}}=18$ kPa. Compute undrained shear strength of each sample and the sensitivity $S_t$. Classify the clay.

$$S_{u,\text{undisturbed}} = \frac{q_u}{2} = \frac{144}{2} = 72\text{ kPa}$$
$$S_{u,\text{remolded}} = \frac{18}{2} = 9\text{ kPa}$$
$$S_t = \frac{S_{u,\text{undisturbed}}}{S_{u,\text{remolded}}} = \frac{72}{9} = 8$$

$S_t=8$ falls in the "sensitive" category ($4 < S_t \leq 8$). Such clays can lose a large fraction of their strength when sampled or remolded, and must be handled carefully in the field.

Answer: $S_u=72$ kPa (undisturbed), $S_u=9$ kPa (remolded), $S_t=8$ — sensitive clay.

Exam Generator Problems

Additional board-style practice items for this topic.

Question Bank: v39

HGE - Geotechnical Engineering / Shear Strength / HGE May 2019

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In a triaxial test on a normally consolidated soil (cohesion $c=0$), failure occurs on a plane where the normal stress is 500 kPa and the shear stress is 280 kPa.

Compute the angle of internal friction, in degrees.

  1. 22.2°
  2. 27.8°
  3. 25.4°
  4. 29.2°

Compute the angle of the failure plane with the major principal plane, in degrees.

  1. 45.3°
  2. 59.6°
  3. 60.8°
  4. 57.2°

Compute the major principal stress at failure, in kPa.

  1. 743 kPa
  2. 919 kPa
  3. 1320 kPa
  4. 978 kPa

Part 1 — Friction angle. The failure envelope passes through the origin, so

$$\tan\phi=\frac{\tau_f}{\sigma_f}.$$

Part 2 — Failure-plane angle.

$$\theta=45^{\circ}+\frac{\phi}{2}.$$

Part 3 — Major principal stress. From the Mohr circle, the radius and center are $R=\dfrac{\tau}{\cos\phi}$ and $\dfrac{R}{\sin\phi}$, so

$$\sigma_1=\frac{R}{\sin\phi}+R.$$

Computed answers:
1. 29.2°
2. 59.6°
3. 978 kPa

Question Bank: v77

HGE - Geotechnical Engineering / Shear Strength / HGE November 2019

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A cohesionless specimen has confining pressure 65 kPa and fails when the additional axial (deviator) stress reaches 150 kPa. Find its internal friction angle.

  1. 24.3 °
  2. 30.1 °
  3. 21.1 °
  4. 32.4 °
The major principal stress is $\sigma_1=\sigma_3+q_d=215$ kPa. For cohesionless soil, $$\phi=\sin^{-1}\left(\frac{\sigma_1-\sigma_3}{\sigma_1+\sigma_3}\right)=32.3923654858^\circ.$$
Computed answer: 32.4 °

Question Bank: v78

HGE - Geotechnical Engineering / Shear Strength / HGE November 2019

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A drained triaxial test has confining pressure 130 kPa and total axial stress at failure 260 kPa. Find the internal friction angle.

  1. 19.5 °
  2. 30.0 °
  3. 32.5 °
  4. 36.0 °
For a cohesionless Mohr-Coulomb envelope, $$\phi=\sin^{-1}\left(\frac{\sigma_1-\sigma_3}{\sigma_1+\sigma_3}\right)=\sin^{-1}\left(\frac{260-130}{260+130}\right)=19.4712206345^\circ.$$
Computed answer: 19.5 °

Question Bank: v79

HGE - Geotechnical Engineering / Shear Strength / HGE November 2019

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A cohesionless soil has confining pressure 65 kPa and maximum principal stress 87.75 kPa in a triaxial test. Evaluate its internal friction angle.

  1. 15.0 °
  2. 8.6 °
  3. 26.0 °
  4. 18.6 °
For cohesionless soil, $$\phi=\sin^{-1}\left(\frac{\sigma_1-\sigma_3}{\sigma_1+\sigma_3}\right)=\sin^{-1}\left(\frac{87.75-65}{87.75+65}\right)=8.56528110755^\circ.$$
Computed answer: 8.6 °

Question Bank: v80

HGE - Geotechnical Engineering / Shear Strength / HGE November 2019

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On the failure plane of a cohesionless soil, normal stress is 130 kPa and shear stress is 60 kPa. Determine the internal friction angle.

  1. 21.1 °
  2. 22.3 °
  3. 28.0 °
  4. 24.8 °
With zero cohesion, the failure envelope is $\tau=\sigma_n\tan\phi$. Therefore $$\phi=\tan^{-1}\left(\frac{\tau}{\sigma_n}\right)=\tan^{-1}\left(\frac{60}{130}\right)=24.7751405688^\circ.$$
Computed answer: 24.8 °

Question Bank: v83

HGE - Geotechnical Engineering / Shear Strength / HGE November 2019

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Evaluate the soil friction angle if the Rankine passive earth-pressure coefficient is 3.75.

  1. 29.0 °
  2. 32.2 °
  3. 35.4 °
  4. 22.6 °
Rankine passive coefficient is $K_p=(1+\sin\phi)/(1-\sin\phi)$. Solving, $$\phi=\sin^{-1}\left(\frac{K_p-1}{K_p+1}\right)=35.3765401519^\circ.$$
Computed answer: 35.4 °

Question Bank: v84

HGE - Geotechnical Engineering / Shear Strength / HGE November 2019

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Evaluate the soil friction angle if the Rankine active earth-pressure coefficient is 0.45.

  1. 22.3 °
  2. 19.4 °
  3. 17.2 °
  4. 25.2 °
Rankine active coefficient is $K_a=(1-\sin\phi)/(1+\sin\phi)$. Hence $$\phi=\sin^{-1}\left(\frac{1-K_a}{1+K_a}\right)=22.2909703748^\circ.$$
Computed answer: 22.3 °

Question Bank: v284

HGE - Geotechnical Engineering / Shear Strength / HGE Refresher Series

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An unconfined compression specimen has diameter 60 mm and fails at load 85 N.

Find undrained cohesion.

  1. 15.03 kPa
  2. 11.74 kPa
  3. 17.09 kPa
  4. 18.95 kPa
For saturated clay in unconfined compression, cu=qu/2.
Computed answers: 15.0313001809.

Question Bank: v326

HGE - Geotechnical Engineering / Shear Strength / HGE Refresher Series

Maximum shear stress acts on a plane at what angle to a principal plane?

  1. 30°
  2. 45°
  3. 60°
  4. 90°
Maximum-shear planes lie 45 degrees from principal planes.

Question Bank: v327

HGE - Geotechnical Engineering / Shear Strength / HGE Refresher Series

In triaxial loading, deviator stress creates shear on which planes?

  1. Horizontal only
  2. Vertical only
  3. Both principal planes
  4. Inclined planes other than the principal horizontal and vertical planes
Principal planes have zero shear.

Question Bank: v328

HGE - Geotechnical Engineering / Shear Strength / HGE Refresher Series

Which statements describe Mohr-Coulomb shear strength?

  1. It rises with normal stress.
  2. It includes cohesion.
  3. It defines the strength envelope.
  4. All of these.
All follow τ=c+σ'tanφ.

Question Bank: v329

HGE - Geotechnical Engineering / Shear Strength / HGE Refresher Series

Which statement about the failure plane is correct?

  1. It always carries maximum shear.
  2. It need not carry maximum shear.
  3. Its shear always equals τmax.
  4. None
Failure occurs where the stress circle touches the strength envelope.

Question Bank: v330

HGE - Geotechnical Engineering / Shear Strength / HGE Refresher Series

Which test is normally performed in the field?

  1. Direct shear
  2. Vane shear
  3. Triaxial
  4. Unconfined compression
The field vane is used for soft clays.

Question Bank: v339

HGE - Geotechnical Engineering / Shear Strength / HGE Refresher Series

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Use the reference SPT-clay correlations qu≈12.5N kPa and cu=qu/2.

Find approximate qu for N1.

  1. 50.0 kPa
  2. 39.1 kPa
  3. 56.9 kPa
  4. 63.1 kPa

Find approximate cu for N2.

  1. 93.8 kPa
  2. 73.2 kPa
  3. 106.6 kPa
  4. 118.2 kPa

Report the SPT value used for consistency classification.

  1. N = 15
  2. N = 12
  3. N = 17
  4. N = 19
Use the stated empirical correlations.
Computed answers: 50, 93.75, 15.

Question Bank: v354

HGE - Geotechnical Engineering / Shear Strength / HGE Refresher Series

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A Mohr circle has center stress 230 kPa and radius 160 kPa.

Find friction angle at tangency.

  1. 0.77 deg
  2. 0.60 deg
  3. 0.87 deg
  4. 0.97 deg

Find shear stress at tangency.

  1. 114.94 kPa
  2. 89.77 kPa
  3. 130.69 kPa
  4. 144.94 kPa

Find normal stress at tangency.

  1. 341.30 kPa
  2. 266.56 kPa
  3. 388.06 kPa
  4. 430.38 kPa
Use Mohr-circle tangency geometry.
Computed answers: 0.769327343523, 114.940602726, 341.304347826.

Question Bank: v358

HGE - Geotechnical Engineering / Shear Strength / HGE Refresher Series

A cohesive soil with unconfined strength 100-200 kPa is classified as what?

  1. Very soft
  2. Soft
  3. Stiff
  4. Hard
That range is stiff.

Question Bank: v359

HGE - Geotechnical Engineering / Shear Strength / HGE Refresher Series

A cohesive soil with unconfined strength 25-50 kPa is classified as what?

  1. Very stiff
  2. Stiff
  3. Hard
  4. Soft
That range is soft.

Question Bank: v360

HGE - Geotechnical Engineering / Shear Strength / HGE Refresher Series

A cohesive soil with unconfined strength 200-400 kPa is classified as what?

  1. Very stiff
  2. Soft
  3. Stiff
  4. Hard
That range is very stiff.

Question Bank: v362

HGE - Geotechnical Engineering / Shear Strength / HGE Refresher Series

At fixed c and φ, what happens to shear strength when normal effective stress decreases?

  1. It increases.
  2. It is unchanged.
  3. It varies inversely.
  4. It decreases.
Mohr-Coulomb strength falls with σ'.

Question Bank: v370

HGE - Geotechnical Engineering / Shear Strength / HGE Refresher Series

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An unconfined compression specimen has cross-sectional area 45 cm2 and fails under an axial load of 5.5 kN.

Find cohesion.

  1. 611.1 kPa
  2. 477.3 kPa
  3. 694.8 kPa
  4. 770.6 kPa

Find unconfined compressive strength.

  1. 1222.2 kPa
  2. 954.6 kPa
  3. 1389.7 kPa
  4. 1541.2 kPa

Find the maximum shear stress (the Mohr-circle radius).

  1. 611.1 kPa
  2. 477.3 kPa
  3. 694.8 kPa
  4. 770.6 kPa
For an unconfined compression test, $\sigma_3=0$, $q_u=P/A$, and the undrained cohesion and maximum shear stress are both $q_u/2$.
Computed answers: 611.111111111, 1222.22222222, and 611.111111111 kPa.

Question Bank: v377

HGE - Geotechnical Engineering / Shear Strength / HGE Refresher Series

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A saturated soil slice on a plane inclined 15° has thickness 7.5 m, unit weights γd=16.5 and γsat=19 kN/m3, and friction angle 26°.

Find dry factor of safety.

  1. -1.38
  2. -0.88
  3. -0.38
  4. 0.12

Find saturated factor of safety.

  1. -0.67
  2. -0.42
  3. -0.17
  4. 0.08

Estimate critical saturated thickness.

  1. -5.00 m
  2. -3.90 m
  3. -5.68 m
  4. -6.30 m
Resolve driving and frictional resistance along the infinite slope.
Computed answers: -1.37705916068, -0.666061772984, -4.99546329738.