Effective stress in soil: practice problems with step-by-step solutions

Free Effective stress in soil practice problems with diagrams, hints and full worked solutions. New values every time, at three levels: water table at surface, water table in the layer, two layers.

Free practice problemEasy
sandγsat​ = 18.5 kN/m³4 mwater tableXz = 3 m

For the soil profile shown, find the pore water pressure u at point X.

z = 3 mzw = 0 mγw = 9.81 kN/m³sand: γsat = 18.5 kN/m³
kPa
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What you practise

  1. Easywater table at surface
  2. Mediumwater table in the layer
  3. Hardtwo layers

Key formulas

Total stress
σ = Σγi·hi
Pore water pressure
u = γw·(z − zw)
Effective stress
σ′ = σ − u
Water
γw = 9.81 kN/m³

Learn Effective stress in soil from scratch

1What effective stress is

Soil is a skeleton of solid grains with small gaps between them, called pores. Below the water table the pores are full of water.

Every point in the ground carries the weight of everything above it. This load per square metre is the total stress σ, in kPa.

Part of that load is carried by the water in the pores, the pore water pressure u. The rest is carried by the grains pressing on each other.

σuuuσ′

The part carried by the grains is the effective stress:

σ′ = σ − u

This is Terzaghi’s principle. Strength and settlement of soil depend on σ′, because only grain-to-grain contact gives friction and compression.

Think of a wet sponge in a bucket of water. Water pressure all around it does not squeeze it. Only a load on the sponge itself does.

Self-check Which stress controls how strong a soil is and how much it settles?
  1. The effective stress σ′
  2. The pore water pressure u
  3. The sum σ + u
  4. The total stress σ

Answer: The effective stress σ′. Only the stress carried grain to grain gives friction and compression. That is σ′ = σ − u. The water carries u, which does not press the grains together.

2Total stress from unit weights

The unit weight γ is the weight of one cubic metre of soil, in kN/m³. A soil band of thickness h adds γ·h to the total stress below it.

σ = Σ γi · hi

Soil above the water table is drained, so you use the dry unit weight γd. Below the water table the pores are full of water, so you use the heavier saturated unit weight γsat.

Typical values: dry sand 15 to 17 kN/m³, saturated sand 18 to 21 kN/m³, water 9.81 kN/m³.

Example: 2 m of dry sand with γd = 16 kN/m³ on top of 3 m of saturated sand with γsat = 20 kN/m³. At 5 m depth σ = 16·2 + 20·3 = 32 + 60 = 92 kPa.

Self-check Dry sand (γd = 16 kN/m³) lies above the water table. What is the total stress 3 m below the surface?
  1. 19 kPa
  2. 29.4 kPa
  3. 48 kPa
  4. 5.3 kPa

Answer: 48 kPa. σ = γ·h = 16 · 3 = 48 kPa. Adding γ and h (19) or dividing (5.3) mixes up the formula, and 29.4 kPa uses the unit weight of water.

3Pore water pressure

Water in the pores is connected, like water in a tank. So the pore water pressure grows with depth below the water table, and is zero at the water table itself.

u = γw · (z − zw)

Here z is the depth of the point, zw the depth of the water table and γw = 9.81 kN/m³. Above the water table u = 0.

Only the depth below the water table counts. The soil type and the layers above do not change u.

Self-check The water table is 2 m below the surface. What is u at a depth of 5 m? (γw = 9.81 kN/m³)
  1. 0 kPa
  2. 19.62 kPa
  3. 29.43 kPa
  4. 49.05 kPa

Answer: 29.43 kPa. The point is 5 − 2 = 3 m below the water table, so u = 9.81 · 3 = 29.43 kPa. Using the full depth of 5 m (49.05 kPa) is the most common slip.

4The method step by step

  1. Mark the point X, its depth z and the depth of the water table zw on the profile.
  2. Split the soil above X into bands: each band has one unit weight. The water table and each layer boundary start a new band.
  3. Total stress: add γ·h for every band. Use γd above the water table and γsat below it.
  4. Pore water pressure: u = γw·(z − zw), or 0 if X is above the water table.
  5. Effective stress: σ′ = σ − u.

Write the units at every step. Unit weights in kN/m³ times depths in m give kPa.

5A worked example

Follow every step. Try to predict the next one before you open it.

Worked example

A sand layer 8 m thick has the water table 2 m below the surface. Find the vertical effective stress σ′ at point X, 6 m deep. Use γw = 9.81 kN/m³.

1

Look at the profile

8 mγd​ = 16 kN/m³γsat​ = 20 kN/m³zw​ = 2 mXz = 6 m

X is at z = 6 m. The water table is at zw = 2 m, so X lies 6 − 2 = 4 m below it.

2

Split into bands

From 0 to 2 m the sand is above the water table: 2 m with γd = 16 kN/m³.

From 2 to 6 m the sand is below it: 4 m with γsat = 20 kN/m³.

3

Total stress

σ = 16 · 2 + 20 · 4 = 32 + 80 = 112 kPa
4

Pore water pressure

u = 9.81 · (6 − 2) = 9.81 · 4 = 39.24 kPa
5

Effective stress

σ′ = σ − u = 112 − 39.24 = 72.76 kPa
6

Check the answer

Below the water table each metre adds γsat − γw = 20 − 9.81 = 10.19 kPa to σ′. So σ′ = 16 · 2 + 10.19 · 4 = 32 + 40.76 = 72.76 kPa. The two routes agree.

Answer

σ′ = 72.76 kPa at X.

6Common mistakes

  • Measuring u from the ground surface. Use only the depth below the water table: z − zw.
  • Using γsat above the water table, or γd below it.
  • Forgetting to subtract u and giving σ when σ′ is asked.
  • In two layers, using one unit weight for both. Each layer has its own γ.
  • Mixing up which value is asked: σ, u or σ′. Read the last line of the question again.
Self-check Sand: γd = 16 kN/m³, γsat = 20 kN/m³, water table at 1 m, γw = 9.81 kN/m³. What is σ′ at 4 m depth?
  1. 36.76 kPa
  2. 46.57 kPa
  3. 50.57 kPa
  4. 76 kPa

Answer: 46.57 kPa. σ = 16·1 + 20·3 = 76 kPa and u = 9.81·3 = 29.43 kPa, so σ′ = 46.57 kPa. Taking u from the surface (9.81·4) gives 36.76 kPa, and 76 kPa forgets u altogether.

7How exam questions look

  • Easy: one saturated sand layer with the water table at the surface. Then σ = γsat·z and u = γw·z.
  • Medium: the water table is inside the layer. Use γd above it and γsat below it, as in the example.
  • Hard: two layers, sand over clay, with X at the middle of the clay. Add a third band with the clay’s own γsat.

The question may ask for σ, for u or for σ′. The method is the same; you stop at the step that gives the asked value.

Learn it step by step in the appLearn it step by step in the app

Worked examples

Example 1Easy
sandγsat​ = 20 kN/m³11 mwater tableXz = 8 m

For the soil profile shown, find the total vertical stress σ at point X.

z = 8 mzw = 0 mγw = 9.81 kN/m³sand: γsat = 20 kN/m³

Find σ (kPa)

Show hints
  1. Effective stress = total stress − pore water pressure.
  2. Use the dry unit weight above the water table and the saturated unit weight below it.
  3. σ = 160 kPa.
Show the answer and the step-by-step solution

Effective stress (Terzaghi): the soil skeleton carries σ′ = σ − u. The total stress σ is the weight of everything above the point, Σγ·h, using the dry unit weight above the water table and the saturated unit weight below it. Below the water table the pore water pressure is hydrostatic: u = γw·(depth below the water table).

Total stress

Add the weight of each soil band above X. Above the water table use the dry unit weight, below it the saturated one.

σ = γsat·z = 20·8 = 160 kPa

Answer

σ = 160 kPa
Example 2Medium
sandγd​ = 16 kN/m³γsat​ = 18.5 kN/m³12 mwater tableXz = 11 m

For the soil profile shown, find the total vertical stress σ at point X.

z = 11 mzw = 1 mγw = 9.81 kN/m³sand: γd = 16 kN/m³sand: γsat = 18.5 kN/m³

Find σ (kPa)

Show hints
  1. Effective stress = total stress − pore water pressure.
  2. Use the dry unit weight above the water table and the saturated unit weight below it.
  3. σ = 201 kPa.
Show the answer and the step-by-step solution

Effective stress (Terzaghi): the soil skeleton carries σ′ = σ − u. The total stress σ is the weight of everything above the point, Σγ·h, using the dry unit weight above the water table and the saturated unit weight below it. Below the water table the pore water pressure is hydrostatic: u = γw·(depth below the water table).

Total stress

Add the weight of each soil band above X. Above the water table use the dry unit weight, below it the saturated one.

σ = γd·zw + γsat·(z − zw) = 16·1 + 18.5·10 = 201 kPa

Answer

σ = 201 kPa
Example 3Hard
sandγd​ = 15 kN/m³γsat​ = 19.5 kN/m³4 mclayγsat​ = 19 kN/m³5 mwater tableXz = 6.5 m

For the soil profile shown, find the vertical effective stress σ′ at point X.

z = 6.5 mzw = 2 mγw = 9.81 kN/m³sand: γd = 15 kN/m³sand: γsat = 19.5 kN/m³clay: γsat = 19 kN/m³

Find σ′ (kPa)

Show hints
  1. Effective stress = total stress − pore water pressure.
  2. Find σ from the weights of the layers above X, and u from the depth of X below the water table.
  3. σ = 116.5 kPa.
Show the answer and the step-by-step solution

Effective stress (Terzaghi): the soil skeleton carries σ′ = σ − u. The total stress σ is the weight of everything above the point, Σγ·h, using the dry unit weight above the water table and the saturated unit weight below it. Below the water table the pore water pressure is hydrostatic: u = γw·(depth below the water table).

Total stress

Add the weight of each soil band above X. Above the water table use the dry unit weight, below it the saturated one.

σ = γd·2 + γsat,sand·2 + γsat,clay·2.5
σ = 15·2 + 19.5·2 + 19·2.5 = 116.5 kPa

Pore water pressure

Water pressure grows with depth below the water table.

u = γw·(z − zw) = 9.81·4.5 = 44.15 kPa

Effective stress

σ′ = σ − u = 116.5 − 44.15 = 72.35 kPa

Answer

σ′ = 72.35 kPa

Questions and answers

What is Effective stress in soil?

Soil is a skeleton of solid grains with small gaps between them, called pores. Below the water table the pores are full of water.

Every point in the ground carries the weight of everything above it. This load per square metre is the total stress σ, in kPa.

How do I solve Effective stress in soil problems step by step?

Effective stress (Terzaghi): the soil skeleton carries σ′ = σ − u. The total stress σ is the weight of everything above the point, Σγ·h, using the dry unit weight above the water table and the saturated unit weight below it. Below the water table the pore water pressure is hydrostatic: u = γw·(depth below the water table).

What do the Effective stress in soil exercises cover?
  • Easy: water table at surface
  • Medium: water table in the layer
  • Hard: two layers

Every problem has new values, so you learn the method instead of an answer. Stuck? Open a small hint first, and the full explanation only when you need it.

How practising works

  • Every problem has new values, so you learn the method instead of an answer.
  • Stuck? Open a small hint first, and the full explanation only when you need it.
  • Your progress, mistakes and saved problems are kept, so you can review before the exam.

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