Geotechnical Engineering: free practice, theory and problems

1. Soil and classification

What is it about?

Everything we build stands on soil or rock. Geotechnical engineering is about how soil behaves when we load it, dig in it or change the groundwater. The first step is always to find out what kind of soil it is, and how much water and air it contains.

Concepts and formulas

• Soil consists of three phases: grains (solids), water and air. The spaces between the grains are called pores.
• Water content: $w = m_w/m_s$ (mass of water divided by mass of dry grains), often in percent.
• Void ratio: $e = V_p/V_s$ (pore volume divided by grain volume).
• Porosity: $n = V_p/V$, and $n = \dfrac{e}{1+e}$.
• Degree of saturation: $S_r = V_w/V_p$. Below the groundwater table the soil is usually fully saturated, $S_r = 1$.
• Unit weight: $\gamma = \rho g$, in kN/m³. Water: $\gamma_w \approx 10$ kN/m³.
• Grain sizes: clay below 0.002 mm, silt 0.002–0.063 mm, sand 0.063–2 mm, gravel 2–63 mm, cobbles above 63 mm.
• Sensitivity: $S_t = s_u/s_r$ (undisturbed strength divided by remoulded strength). Quick clay has a very high sensitivity and turns liquid when it is disturbed.

How to solve the problems

1. Write down what is known: masses, volumes or ratios.
2. Draw a phase diagram: grains at the bottom, then water, then air.
3. Use the definition directly. Watch what is in the numerator and the denominator.

Example

A sample weighs 190 g wet and 160 g after drying. What is the water content?

1. Mass of water: $190 - 160 = 30$ g.
2. $w = 30/160 = 0.1875 \approx 19$ %.

Common mistakes

• Dividing by the wet mass. Water content is always relative to the dry mass.
• Mixing up void ratio $e$ and porosity $n$. The void ratio can exceed 1, the porosity cannot.
• Thinking clay and silt are the same. Clay has much smaller grains and behaves very differently.

Concepts in this part

Practise soil and classification in the app →

2. Effective stress and settlement

What is it about?

The weight of the soil above a point gives a stress. But the water in the pores is under pressure and carries part of the load. Only the rest, the effective stress, pushes the grains together and gives the soil its strength. This may be the most important idea in all of geotechnics.

Concepts and formulas

• Total stress (vertical): $\sigma_v = \sum \gamma_i h_i$, the sum of unit weight times thickness for the layers above.
• Pore pressure (hydrostatic): $u = \gamma_w h_w$, where $h_w$ is the depth below the groundwater table and $\gamma_w \approx 10$ kN/m³.
• Effective stress: $\sigma' = \sigma - u$ (Terzaghi's principle).
• Below the groundwater table you can also use the submerged unit weight $\gamma' = \gamma - \gamma_w$ directly.
• Settlement of a layer of thickness $H$ and oedometer modulus $M$: $s = \dfrac{\Delta\sigma'\, H}{M}$.
• If the groundwater is lowered, $u$ goes down and $\sigma'$ goes up. Clay under buildings can then settle.

How to solve the problems

1. Draw a column with the layers and the groundwater table.
2. Compute $\sigma_v$ by summing layer by layer down to the point.
3. Compute $u$ from the depth below the groundwater table.
4. Subtract: $\sigma' = \sigma_v - u$.

Example

Sand with $\gamma = 20$ kN/m³, groundwater 2 m below ground. Find $\sigma'$ at 6 m depth.

1. $\sigma_v = 20\cdot 6 = 120$ kPa.
2. $u = 10\cdot (6-2) = 40$ kPa.
3. $\sigma' = 120 - 40 = 80$ kPa.

Common mistakes

• Computing pore pressure from the ground surface instead of from the groundwater table.
• Using $\gamma'$ and also subtracting $u$. That removes the water twice.
• Forgetting that settlement comes from the change in effective stress, not total stress.

Concepts in this part

Practise effective stress and settlement in the app →

3. Earth pressure, bearing capacity and stability

What is it about?

A retaining wall must withstand the pressure from the soil behind it, a foundation must not sink through the ground, and a slope must not slide. All three are about the same thing: the soil's shear strength against the forces trying to make it slide.

Concepts and formulas

• The friction angle $\varphi$ describes the strength of sand and gravel (drained). The undrained shear strength $s_u$ describes clay under fast loading.
• Active earth pressure (the soil pushes on a wall that yields slightly):

• Passive earth pressure (the wall is pushed into the soil): $K_p = 1/K_a$.
• Force per metre of wall (dry soil, height $H$): $P_a = \tfrac{1}{2} K_a \gamma H^2$, acting at height $H/3$ above the base.
• Bearing capacity, strip footing on clay (undrained): $q_u \approx 5.14\, s_u$ (plus any overburden pressure).
• Factor of safety: $F = \dfrac{\text{resistance}}{\text{driving force}}$. For an infinitely long, dry sand slope: $F = \dfrac{\tan\varphi}{\tan\beta}$.

How to solve the problems

1. Decide whether it is sand ($\varphi$) or clay ($s_u$), and whether the soil pushes (active) or is pushed (passive).
2. Compute the earth pressure coefficient or bearing capacity factor.
3. Insert into the formula and check the units (kPa = kN/m², kN per metre of wall).

Example

Sand with $\varphi = 30^\circ$ and $\gamma = 18$ kN/m³ behind a 4 m high wall. Find the active force.

1. $K_a = (1 - \sin 30^\circ)/(1 + \sin 30^\circ) = 0.5/1.5 = 1/3$.
2. $P_a = \tfrac12 \cdot \tfrac13 \cdot 18 \cdot 4^2 = 48$ kN/m.

Common mistakes

• Swapping active and passive. Passive pressure is always much larger than active.
• Forgetting the square in $H^2$. Double the height gives four times the force.
• Using $\varphi$ for clay under fast loading. Then $s_u$ applies.

Concepts in this part

Practise earth pressure, bearing capacity and stability in the app →