Shear strength of soil
Chapter 8 - a very high probability topic in Paper II. Sorted by exam weightage, with full answers.
Introduction
Every soil supports load only because it has shear strength. If shear stress exceeds shear strength, the soil fails.
Real-life example
Push a heavy cupboard: at first it does not move, but past a certain force it suddenly slides. The same thing happens inside soil - shear stress builds up until it exceeds the soil's shear strength, and failure occurs.
Definition
Shear strength is the maximum resistance offered by soil against sliding along a failure plane.
Why it matters
Shear strength controls bearing capacity, slope stability, earth pressure, retaining walls, foundations, embankments, dams and pavements.
Components of shear strength
Shear strength comes from two sources - cohesion and internal friction.
Cohesion (c)
Internal bonding between soil particles. Common in clay and cemented soils; not significant in clean dry sand. Unit: kPa.
Example: wet clay can be moulded because particles stick together - that sticking force is cohesion.
Internal friction (phi)
Resistance from friction between particles. Mostly important in sand and gravel.
Example: walking on dry sand is difficult because particles slide against each other - that resistance is internal friction.
Memory trick
Clay leads with cohesion. Sand leads with friction.
Shear strength equation - Mohr-Coulomb theory
This is one of the most important equations in geotechnical engineering.
Effective stress up
Shear strength increases.
Cohesion up
Shear strength increases.
Phi up
Shear strength increases.
Types of soil by shear strength
Sand, gravel
c = 0. Strength depends only on phi.
Clay
Both c and phi contribute to strength.
Principal stresses and principal plane
At every point in soil there are three principal stresses; for most geotechnical problems only sigma1 (largest) and sigma3 (smallest) matter.
sigma1
Largest principal stress.
sigma2
Intermediate principal stress.
sigma3
Smallest principal stress.
Principal plane
A plane on which shear stress is zero - only normal stress acts on it. This concept is essential for understanding the Mohr circle.
Mohr circle
A graphical method to determine normal stress, shear stress, principal stresses, principal planes and the failure condition.
Radius = (sigma1 - sigma3) / 2
tau max = (sigma1 - sigma3) / 2
Mohr-Coulomb failure theory
Failure occurs the moment the Mohr circle becomes tangent to the failure envelope.
Intercept
Equals cohesion (c).
Slope
Equals tan(phi).
Interpretation
If c increases, the failure envelope shifts upward. If phi increases, the envelope becomes steeper. Both increases mean higher soil strength.
The four shear tests
Best for sand and gravel
Apparatus: shear box, loading frame, dial gauges.
Procedure: place specimen, apply normal load, apply horizontal shear load, record failure load, plot graph.
Advantages: simple, cheap, fast.
Limitations: failure plane is predetermined; not suitable for all soils.
Best for saturated clay
No lateral pressure applied - a very simple test.
P = maximum load, A = corrected area.
Very important MCQ formula.
The most reliable lab test
Most accurate; drainage can be controlled; failure plane develops naturally; suitable for almost all soils.
- UU - unconsolidated undrained - fastest test
- CU - consolidated undrained - measures pore pressure, most common in practice
- CD - consolidated drained - very slow, best for long-term conditions
Best for very soft clay
A field test. Consists of four blades rotated inside the clay.
Advantages: simple, fast, gives undisturbed strength.
Memory trick - triaxial speed
UU is quick, CU is medium, CD is slow.
Comparison of shear tests
| Test | Best soil | Accuracy |
|---|---|---|
| Direct Shear | Sand | Medium |
| UCS | Clay | Medium |
| Triaxial | All soils | High |
| Vane Shear | Soft clay | High (field) |
Engineering applications: foundation design, retaining walls, earth dams, slopes, pavements, tunnel stability, embankments.
Questions by weightage
Combines the syllabus question bank with the commonly repeated Loksewa question set. Highest-value questions first.
| Marks | Question | Priority |
|---|---|---|
| 10 | Explain Mohr-Coulomb failure theory with a neat sketch | high |
| 10 | Explain the Triaxial Test with its types (UU, CU, CD) | high |
| 10 | Explain the Mohr circle and derive the failure condition | high |
| 5 | Define shear strength | medium |
| 5 | Explain cohesion and angle of internal friction | medium |
| 5 | Explain the Direct Shear Test | medium |
| 5 | Explain the Vane Shear Test | medium |
| 5 | Explain the UCS Test | medium |
| 5 | Compare / differentiate Direct Shear Test and Triaxial Test | medium |
| 1 each | Objective / MCQ set (4 questions below) | scoring |
Question 1 - Explain Mohr-Coulomb failure theory with a neat sketch
The Mohr-Coulomb failure theory states that a soil fails at a point when the shear stress on some plane through that point reaches a critical combination of cohesion and friction. This condition is expressed as the failure envelope:
tau = c + sigma tan(phi)
where tau is the shear stress on the failure plane, sigma is the normal stress on that plane, c is the cohesion intercept and phi is the angle of internal friction. Plotted on a graph of shear stress (tau) against normal stress (sigma), the failure envelope is a straight line whose intercept on the tau axis is c and whose slope is tan(phi).
Sketch description
Draw sigma on the horizontal axis and tau on the vertical axis. The failure envelope is a straight line starting at height c on the tau axis and rising at angle phi. The stress state at a point is represented by a Mohr circle, with its centre at (sigma1 + sigma3)/2 on the sigma axis and radius (sigma1 - sigma3)/2, where sigma1 and sigma3 are the major and minor principal stresses.
Failure condition
- If the Mohr circle lies entirely below the failure envelope, the soil is safe - stresses on every plane are below the strength available
- If the circle is tangent to the envelope, the soil is at the point of failure, and the point of tangency gives the orientation of the actual failure plane
- A circle that crosses the envelope is not physically possible, since failure would already have occurred before that stress state was reached
Interpretation
An increase in cohesion shifts the envelope upward; an increase in the friction angle makes the envelope steeper. Both changes increase the strength of the soil and therefore the size of Mohr circle it can safely support before failure.
Question 2 - Explain the Triaxial Test and its types
The triaxial compression test is the most reliable laboratory method for determining the shear strength parameters of soil. A cylindrical soil specimen, enclosed in a rubber membrane, is placed inside a pressure cell. An all-round confining pressure is applied through the cell fluid, and an additional axial load is applied through a loading ram until the specimen fails in shear.
Advantages
- Most accurate of the standard shear tests
- Drainage conditions can be controlled during the test
- The failure plane develops naturally, rather than being fixed in advance
- Suitable for almost all types of soil
Types of triaxial test
- UU (unconsolidated undrained): no drainage is allowed during either the application of confining pressure or the axial load. It is the fastest of the three tests and gives the undrained shear strength, useful for short-term stability of saturated clay.
- CU (consolidated undrained): the specimen is first allowed to consolidate under the confining pressure with drainage open, then sheared with drainage closed while pore pressure is measured. It is a medium-duration test and is the most common type used in practice, since it gives both total and effective stress parameters.
- CD (consolidated drained): the specimen is consolidated and then sheared very slowly with drainage open throughout, so that no excess pore pressure develops. It is the slowest test but gives the effective stress parameters directly, and is best for assessing long-term stability.
Memory trick
UU is quick, CU is medium, CD is slow - the drainage conditions allowed during shearing determine both the test duration and which strength parameters are obtained.
Question 3 - Explain the Mohr circle and derive the failure condition
The Mohr circle is a graphical construction that represents the state of stress at a point in a soil mass. Given the major principal stress sigma1 and the minor principal stress sigma3 acting on a point, a circle is drawn on axes of normal stress (sigma, horizontal) and shear stress (tau, vertical):
Centre = (sigma1 + sigma3) / 2 Radius = (sigma1 - sigma3) / 2
Every point on the circle represents the normal and shear stress acting on a plane through that point, at some particular orientation. The topmost point of the circle gives the maximum shear stress:
tau max = (sigma1 - sigma3) / 2
Deriving the failure condition
The Mohr-Coulomb failure envelope, tau = c + sigma tan(phi), is plotted on the same axes. Failure occurs at the point where the Mohr circle just touches (is tangent to) this envelope. At that point of tangency, using the geometry of the circle and the envelope line, it can be shown that the relationship between the principal stresses at failure is:
sigma1 = sigma3 tan^2(45 + phi/2) + 2c tan(45 + phi/2)
This equation gives the major principal stress at which failure occurs for a given minor principal stress, cohesion and friction angle, and is derived directly from the tangency condition between the Mohr circle and the Mohr-Coulomb failure envelope.
Question 4 - Define shear strength
Shear strength is the maximum resistance offered by soil against sliding along a failure plane. It is the internal resistance per unit area that a soil mass can offer to resist failure along any plane, and it is what allows soil to support foundations, retain earth pressure and maintain stable slopes. According to the Mohr-Coulomb theory, shear strength is expressed as tau = c + sigma' tan(phi), made up of a cohesive component (c) and a frictional component (sigma' tan phi). If the shear stress induced by external loading exceeds this available shear strength, the soil fails along that plane.
Question 5 - Explain cohesion and angle of internal friction
Cohesion (c) is the internal bonding between soil particles, arising from electrochemical attraction and cementation between fine particles. It is significant in clay and cemented soils, but negligible in clean, dry sand. It is measured in kPa. Wet clay can be moulded into a shape without falling apart because its particles stick together - that sticking force is cohesion.
Angle of internal friction (phi) is a measure of the frictional resistance developed between soil particles as they slide or roll over one another. It is most significant in sand and gravel, where particles interlock and resist sliding. This is why walking on dry, loose sand is difficult - the particles slide against each other and generate frictional resistance, exactly the mechanism captured by phi.
Together, c and phi define the two components of the Mohr-Coulomb shear strength equation, tau = c + sigma' tan(phi).
Question 6 - Explain the Direct Shear Test
The direct shear test is a laboratory test used to determine the cohesion and angle of friction of a soil, and is most suitable for sand and gravel. The apparatus consists of a split shear box, a loading frame and dial gauges to measure deformation.
Procedure
- Place the soil specimen in the two-part shear box
- Apply a normal (vertical) load on top of the specimen
- Apply a horizontal shear load, increasing it gradually until the specimen fails along the horizontal plane between the two halves of the box
- Record the shear load at failure
- Repeat at different normal loads and plot shear stress at failure against normal stress - the resulting line gives cohesion (intercept) and the friction angle (slope)
Advantages
Simple, cheap and fast to perform.
Limitations
The failure plane is predetermined by the shear box rather than developing naturally, and the test is not suitable for all soil types, particularly saturated clays where drainage cannot be controlled.
Question 7 - Explain the Vane Shear Test
The vane shear test is a field test used to determine the undrained shear strength of very soft, saturated clay in situ. It consists of a rod with four thin metal blades (a vane) that is pushed into the clay and then rotated at a constant rate. The torque required to shear the soil along the cylindrical surface traced by the vane is measured, and this torque is converted into the undrained shear strength using the geometry of the vane.
Advantages
Simple and fast to carry out, and because it is performed in situ, it gives the undisturbed shear strength of the soil without the sampling disturbance that affects laboratory tests.
Question 8 - Explain the UCS Test
The Unconfined Compression Test (UCS) is a simple laboratory test suitable for saturated clay. A cylindrical clay specimen is loaded axially with no lateral (confining) pressure applied, and the load is increased until the specimen fails. Because it applies no confining pressure, it is essentially a special case of the triaxial test with zero cell pressure.
qu = P / A
where P is the maximum load at failure and A is the corrected cross-sectional area of the specimen. The undrained cohesion is then obtained from:
cu = qu / 2
This relationship, cu = qu / 2, is one of the most frequently tested formulas in the Loksewa objective section.
Question 9 - Compare Direct Shear Test and Triaxial Test
| Basis | Direct Shear Test | Triaxial Test |
|---|---|---|
| Failure plane | Predetermined by the shear box | Develops naturally at the weakest orientation |
| Drainage control | Cannot be controlled | Can be fully controlled (UU, CU, CD) |
| Stress state | Stress distribution on the failure plane is non-uniform | Stress state is well-defined and uniform |
| Suitable soils | Best for sand and gravel | Suitable for almost all soil types |
| Accuracy | Medium | High - the most reliable laboratory test |
| Cost and speed | Simple, cheap, fast | More equipment and time required, but far more versatile |
MCQs
Formula to memorize
Undrained cohesion: cu = qu / 2 - this exact relationship is frequently tested directly as an objective question.
Likely interview questions
- Why is the Triaxial Test considered superior to the Direct Shear Test?
- Why is UCS unsuitable for sand?
- What is the physical meaning of cohesion?
- Why does effective stress control shear strength?
- When would you choose a Vane Shear Test over a laboratory test?
Memory box
- Shear strength = resistance against sliding.
- Mohr-Coulomb equation: tau = c + sigma' tan(phi).
- Clay - cohesion dominates. Sand - friction dominates.
- Mohr circle graphically represents the stress state and the failure condition.
- UCS: cu = qu / 2.
- Direct Shear: best for sands.
- Triaxial Test: most versatile and reliable.
- Vane Shear Test: best for soft saturated clays.
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