Vertical Stress Profiles with Surcharge

Example 1

Soil phase relationships from field data

A sample has moist unit weight 19.2 kN/m³, water content 16% and Gs = 2.68. Find the dry unit weight, void ratio and degree of saturation.

Given

$\gamma = 19.2 kN/m^{3}$
$w = 0.16$
$Gs = 2.68$
γw = 9.81 kN/m³

Find

γ_d, e, S

Start with the thinking

Dry unit weight first — everything else follows.
Se = wGs closes the saturation calculation.

Figure 1 — schematic for Soil phase relationships from field data

Step-by-step solution

Dry unit weight
$\gamma_d = \gamma/(1 + w) = 19.2/1.16 = 16.55 kN/m^{3}$
Void ratio — e = Gsγw/γ_d − 1 = 2.68(9.81)/16.55 − 1
Evaluate
$e = 26.29/16.55 - 1 = 1.589 - 1 = 0.589$
Saturation
$S = wGs/e = 0.16(2.68)/0.589$
Result
$\gamma_d = 16.6 kN/m^{3}, e = 0.589, S = 0.728 (72.8%)$

Answer:

\gamma_d = 16.6 kN/m^{3}, e = 0.589, S = 72.8%

Why the other options are there

γ_d = 22.3 kN/m³ (multiplied by 1 + w)
S = 100% (assumed saturated)

Reference: FE Reference Handbook — Geotechnical — Phase relationships

Example 2

Effective stress below a water table

A profile has 3 m of moist sand (γ = 18.0 kN/m³) over saturated sand (γ_sat = 20.5 kN/m³) with the water table at 3 m. Find the effective stress at 8 m depth.

Given

0–3 m:
$\gamma = 18.0 kN/m^{3}$
3–8 m:
$\gamma_sat = 20.5 kN/m^{3}$
Water table at 3 m

Find

σ′ at 8 m

Start with the thinking

Total stress accumulates layer by layer.
Pore pressure is measured from the water table, not the ground surface.

Step-by-step solution

Total stress
$\sigma = 18.0(3) + 20.5(5) = 54.0 + 102.5 = 156.5 kPa$
Pore pressure — u = γw(8 − 3) = 9.81(5) = 49.1 kPa
Effective stress
$\sigma' = \sigma - u = 156.5 - 49.1$
Result
$\sigma' = 107.4 kPa$

Answer:

\sigma' = 107 kPa

Why the other options are there

156.5 kPa (pore pressure ignored)
78.0 kPa (u measured from the surface)

Reference: FE Reference Handbook — Geotechnical — Effective stress

Example 3

Vertical stress profile with surcharge and the shear stress at failure — Vertical Stress Profiles with Surcharge

A profile has 2.5 m of moist sand (γ = 20.0 kN/m³) over 8.5 m of saturated sand (γ_sat = 20.0 kN/m³) with the water table at the interface. A surcharge of 23 kPa acts at the surface. At the base compute the total normal stress, pore pressure, effective stress, horizontal stress (K₀ = 0.40) and the shear stress at failure for φ′ = 36°.

Given

Layer 1:
$2.5 m, \gamma = 20.0 kN/m^{3}$
Layer 2:
$8.5 m, \gamma_sat = 20.0 kN/m^{3}$
$q = 23 kPa$
$K_{0} = 0.40, \phi' = 36^{\circ}$

Find

σ_v, u, σ′_v, σ′_h and τ_f at the base

Start with the thinking

Total normal stress accumulates every overburden layer plus any surcharge.
Only effective stress governs strength — the horizontal stress and shear stress at failure both use σ′.

Step-by-step solution

Formula — σ_v = q + Σγh
Substituting
$\sigma_v = 23 + 20.0(2.5) + 20.0(8.5) = 243.0 kPa$
Formula
$u = \gamma_w h_w$
Substituting
$u = 9.81(8.5) = 83.39 kPa$
Formula
$\sigma'_v = \sigma_v - u$
Substituting
$\sigma'_v = 243.0 - 83.39 = 159.6 kPa$
Formula
$\sigma'_h = K_{0} \sigma'_v$
Substituting
$\sigma'_h = 0.40(159.6) = 63.85 kPa$
Formula
$\tau_f = \sigma' \tan \phi'$
Substituting
$\tau_f = 159.6 \tan 36^{\circ} = 116.0 kPa$

Answer:

\sigma_v = 243.0 kPa, u = 83.4 kPa, \sigma'_v = 159.6 kPa, \sigma'_h = 63.8 kPa, \tau_f = 116.0 kPa

Why the other options are there

τ_f = 176.5 kPa (total stress used)
σ′_h = 97.2 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Vertical Stress Profiles with Surcharge

Example 4

Vertical stress profile with surcharge and the shear stress at failure — Vertical Stress Profiles with Surcharge (2)

A profile has 4.5 m of moist sand (γ = 18.5 kN/m³) over 7.0 m of saturated sand (γ_sat = 19.5 kN/m³) with the water table at the interface. A surcharge of 25 kPa acts at the surface. At the base compute the total normal stress, pore pressure, effective stress, horizontal stress (K₀ = 0.45) and the shear stress at failure for φ′ = 35°.

Given

Layer 1:
$4.5 m, \gamma = 18.5 kN/m^{3}$
Layer 2:
$7.0 m, \gamma_sat = 19.5 kN/m^{3}$
$q = 25 kPa$
$K_{0} = 0.45, \phi' = 35^{\circ}$

Find

σ_v, u, σ′_v, σ′_h and τ_f at the base

Start with the thinking

Total normal stress accumulates every overburden layer plus any surcharge.
Only effective stress governs strength — the horizontal stress and shear stress at failure both use σ′.

Step-by-step solution

Formula — σ_v = q + Σγh
Substituting
$\sigma_v = 25 + 18.5(4.5) + 19.5(7.0) = 244.8 kPa$
Formula
$u = \gamma_w h_w$
Substituting
$u = 9.81(7.0) = 68.67 kPa$
Formula
$\sigma'_v = \sigma_v - u$
Substituting
$\sigma'_v = 244.8 - 68.67 = 176.1 kPa$
Formula
$\sigma'_h = K_{0} \sigma'_v$
Substituting
$\sigma'_h = 0.45(176.1) = 79.24 kPa$
Formula
$\tau_f = \sigma' \tan \phi'$
Substituting
$\tau_f = 176.1 \tan 35^{\circ} = 123.3 kPa$

Answer:

\sigma_v = 244.8 kPa, u = 68.7 kPa, \sigma'_v = 176.1 kPa, \sigma'_h = 79.2 kPa, \tau_f = 123.3 kPa

Why the other options are there

τ_f = 171.4 kPa (total stress used)
σ′_h = 110.1 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Vertical Stress Profiles with Surcharge

Example 5

Vertical stress profile with surcharge and the shear stress at failure — Vertical Stress Profiles with Surcharge (3)

A profile has 3.0 m of moist sand (γ = 18.5 kN/m³) over 5.0 m of saturated sand (γ_sat = 19.5 kN/m³) with the water table at the interface. A surcharge of 50 kPa acts at the surface. At the base compute the total normal stress, pore pressure, effective stress, horizontal stress (K₀ = 0.45) and the shear stress at failure for φ′ = 27°.

Given

Layer 1:
$3.0 m, \gamma = 18.5 kN/m^{3}$
Layer 2:
$5.0 m, \gamma_sat = 19.5 kN/m^{3}$
$q = 50 kPa$
$K_{0} = 0.45, \phi' = 27^{\circ}$

Find

σ_v, u, σ′_v, σ′_h and τ_f at the base

Start with the thinking

Total normal stress accumulates every overburden layer plus any surcharge.
Only effective stress governs strength — the horizontal stress and shear stress at failure both use σ′.

Step-by-step solution

Formula — σ_v = q + Σγh
Substituting
$\sigma_v = 50 + 18.5(3.0) + 19.5(5.0) = 203.0 kPa$
Formula
$u = \gamma_w h_w$
Substituting
$u = 9.81(5.0) = 49.05 kPa$
Formula
$\sigma'_v = \sigma_v - u$
Substituting
$\sigma'_v = 203.0 - 49.05 = 154.0 kPa$
Formula
$\sigma'_h = K_{0} \sigma'_v$
Substituting
$\sigma'_h = 0.45(154.0) = 69.28 kPa$
Formula
$\tau_f = \sigma' \tan \phi'$
Substituting
$\tau_f = 154.0 \tan 27^{\circ} = 78.44 kPa$

Answer:

\sigma_v = 203.0 kPa, u = 49.1 kPa, \sigma'_v = 154.0 kPa, \sigma'_h = 69.3 kPa, \tau_f = 78.4 kPa

Why the other options are there

τ_f = 103.4 kPa (total stress used)
σ′_h = 91.4 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Vertical Stress Profiles with Surcharge

Example 6

Vertical stress profile with surcharge and the shear stress at failure — Vertical Stress Profiles with Surcharge (4)

A profile has 5.0 m of moist sand (γ = 17.0 kN/m³) over 7.5 m of saturated sand (γ_sat = 21.5 kN/m³) with the water table at the interface. A surcharge of 9 kPa acts at the surface. At the base compute the total normal stress, pore pressure, effective stress, horizontal stress (K₀ = 0.45) and the shear stress at failure for φ′ = 32°.

Given

Layer 1:
$5.0 m, \gamma = 17.0 kN/m^{3}$
Layer 2:
$7.5 m, \gamma_sat = 21.5 kN/m^{3}$
$q = 9 kPa$
$K_{0} = 0.45, \phi' = 32^{\circ}$

Find

σ_v, u, σ′_v, σ′_h and τ_f at the base

Start with the thinking

Total normal stress accumulates every overburden layer plus any surcharge.
Only effective stress governs strength — the horizontal stress and shear stress at failure both use σ′.

Step-by-step solution

Formula — σ_v = q + Σγh
Substituting
$\sigma_v = 9 + 17.0(5.0) + 21.5(7.5) = 255.3 kPa$
Formula
$u = \gamma_w h_w$
Substituting
$u = 9.81(7.5) = 73.58 kPa$
Formula
$\sigma'_v = \sigma_v - u$
Substituting
$\sigma'_v = 255.3 - 73.58 = 181.7 kPa$
Formula
$\sigma'_h = K_{0} \sigma'_v$
Substituting
$\sigma'_h = 0.45(181.7) = 81.75 kPa$
Formula
$\tau_f = \sigma' \tan \phi'$
Substituting
$\tau_f = 181.7 \tan 32^{\circ} = 113.5 kPa$

Answer:

\sigma_v = 255.3 kPa, u = 73.6 kPa, \sigma'_v = 181.7 kPa, \sigma'_h = 81.8 kPa, \tau_f = 113.5 kPa

Why the other options are there

τ_f = 159.5 kPa (total stress used)
σ′_h = 114.9 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Vertical Stress Profiles with Surcharge

Example 7

Vertical stress profile with surcharge and the shear stress at failure — Vertical Stress Profiles with Surcharge (5)

A profile has 5.0 m of moist sand (γ = 18.5 kN/m³) over 6.5 m of saturated sand (γ_sat = 21.5 kN/m³) with the water table at the interface. A surcharge of 46 kPa acts at the surface. At the base compute the total normal stress, pore pressure, effective stress, horizontal stress (K₀ = 0.60) and the shear stress at failure for φ′ = 30°.

Given

Layer 1:
$5.0 m, \gamma = 18.5 kN/m^{3}$
Layer 2:
$6.5 m, \gamma_sat = 21.5 kN/m^{3}$
$q = 46 kPa$
$K_{0} = 0.60, \phi' = 30^{\circ}$

Find

σ_v, u, σ′_v, σ′_h and τ_f at the base

Start with the thinking

Total normal stress accumulates every overburden layer plus any surcharge.
Only effective stress governs strength — the horizontal stress and shear stress at failure both use σ′.

Step-by-step solution

Formula — σ_v = q + Σγh
Substituting
$\sigma_v = 46 + 18.5(5.0) + 21.5(6.5) = 278.3 kPa$
Formula
$u = \gamma_w h_w$
Substituting
$u = 9.81(6.5) = 63.77 kPa$
Formula
$\sigma'_v = \sigma_v - u$
Substituting
$\sigma'_v = 278.3 - 63.77 = 214.5 kPa$
Formula
$\sigma'_h = K_{0} \sigma'_v$
Substituting
$\sigma'_h = 0.60(214.5) = 128.7 kPa$
Formula
$\tau_f = \sigma' \tan \phi'$
Substituting
$\tau_f = 214.5 \tan 30^{\circ} = 123.8 kPa$

Answer:

\sigma_v = 278.3 kPa, u = 63.8 kPa, \sigma'_v = 214.5 kPa, \sigma'_h = 128.7 kPa, \tau_f = 123.8 kPa

Why the other options are there

τ_f = 160.6 kPa (total stress used)
σ′_h = 167.0 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Vertical Stress Profiles with Surcharge

Example 8

Vertical stress profile with surcharge and the shear stress at failure — Vertical Stress Profiles with Surcharge (6)

A profile has 4.0 m of moist sand (γ = 19.5 kN/m³) over 6.0 m of saturated sand (γ_sat = 19.0 kN/m³) with the water table at the interface. A surcharge of 6 kPa acts at the surface. At the base compute the total normal stress, pore pressure, effective stress, horizontal stress (K₀ = 0.55) and the shear stress at failure for φ′ = 27°.

Given

Layer 1:
$4.0 m, \gamma = 19.5 kN/m^{3}$
Layer 2:
$6.0 m, \gamma_sat = 19.0 kN/m^{3}$
$q = 6 kPa$
$K_{0} = 0.55, \phi' = 27^{\circ}$

Find

σ_v, u, σ′_v, σ′_h and τ_f at the base

Start with the thinking

Total normal stress accumulates every overburden layer plus any surcharge.
Only effective stress governs strength — the horizontal stress and shear stress at failure both use σ′.

Step-by-step solution

Formula — σ_v = q + Σγh
Substituting
$\sigma_v = 6 + 19.5(4.0) + 19.0(6.0) = 198.0 kPa$
Formula
$u = \gamma_w h_w$
Substituting
$u = 9.81(6.0) = 58.86 kPa$
Formula
$\sigma'_v = \sigma_v - u$
Substituting
$\sigma'_v = 198.0 - 58.86 = 139.1 kPa$
Formula
$\sigma'_h = K_{0} \sigma'_v$
Substituting
$\sigma'_h = 0.55(139.1) = 76.53 kPa$
Formula
$\tau_f = \sigma' \tan \phi'$
Substituting
$\tau_f = 139.1 \tan 27^{\circ} = 70.90 kPa$

Answer:

\sigma_v = 198.0 kPa, u = 58.9 kPa, \sigma'_v = 139.1 kPa, \sigma'_h = 76.5 kPa, \tau_f = 70.9 kPa

Why the other options are there

τ_f = 100.9 kPa (total stress used)
σ′_h = 108.9 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Vertical Stress Profiles with Surcharge

Example 9

Vertical stress profile with surcharge and the shear stress at failure — Vertical Stress Profiles with Surcharge (7)

A profile has 5.0 m of moist sand (γ = 19.5 kN/m³) over 6.5 m of saturated sand (γ_sat = 20.5 kN/m³) with the water table at the interface. A surcharge of 40 kPa acts at the surface. At the base compute the total normal stress, pore pressure, effective stress, horizontal stress (K₀ = 0.40) and the shear stress at failure for φ′ = 29°.

Given

Layer 1:
$5.0 m, \gamma = 19.5 kN/m^{3}$
Layer 2:
$6.5 m, \gamma_sat = 20.5 kN/m^{3}$
$q = 40 kPa$
$K_{0} = 0.40, \phi' = 29^{\circ}$

Find

σ_v, u, σ′_v, σ′_h and τ_f at the base

Start with the thinking

Total normal stress accumulates every overburden layer plus any surcharge.
Only effective stress governs strength — the horizontal stress and shear stress at failure both use σ′.

Step-by-step solution

Formula — σ_v = q + Σγh
Substituting
$\sigma_v = 40 + 19.5(5.0) + 20.5(6.5) = 270.8 kPa$
Formula
$u = \gamma_w h_w$
Substituting
$u = 9.81(6.5) = 63.77 kPa$
Formula
$\sigma'_v = \sigma_v - u$
Substituting
$\sigma'_v = 270.8 - 63.77 = 207.0 kPa$
Formula
$\sigma'_h = K_{0} \sigma'_v$
Substituting
$\sigma'_h = 0.40(207.0) = 82.79 kPa$
Formula
$\tau_f = \sigma' \tan \phi'$
Substituting
$\tau_f = 207.0 \tan 29^{\circ} = 114.7 kPa$

Answer:

\sigma_v = 270.8 kPa, u = 63.8 kPa, \sigma'_v = 207.0 kPa, \sigma'_h = 82.8 kPa, \tau_f = 114.7 kPa

Why the other options are there

τ_f = 150.1 kPa (total stress used)
σ′_h = 108.3 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Vertical Stress Profiles with Surcharge

Example 10

Vertical stress profile with surcharge and the shear stress at failure — Vertical Stress Profiles with Surcharge (8)

A profile has 3.0 m of moist sand (γ = 17.5 kN/m³) over 6.5 m of saturated sand (γ_sat = 21.0 kN/m³) with the water table at the interface. A surcharge of 19 kPa acts at the surface. At the base compute the total normal stress, pore pressure, effective stress, horizontal stress (K₀ = 0.45) and the shear stress at failure for φ′ = 33°.

Given

Layer 1:
$3.0 m, \gamma = 17.5 kN/m^{3}$
Layer 2:
$6.5 m, \gamma_sat = 21.0 kN/m^{3}$
$q = 19 kPa$
$K_{0} = 0.45, \phi' = 33^{\circ}$

Find

σ_v, u, σ′_v, σ′_h and τ_f at the base

Start with the thinking

Total normal stress accumulates every overburden layer plus any surcharge.
Only effective stress governs strength — the horizontal stress and shear stress at failure both use σ′.

Step-by-step solution

Formula — σ_v = q + Σγh
Substituting
$\sigma_v = 19 + 17.5(3.0) + 21.0(6.5) = 208.0 kPa$
Formula
$u = \gamma_w h_w$
Substituting
$u = 9.81(6.5) = 63.77 kPa$
Formula
$\sigma'_v = \sigma_v - u$
Substituting
$\sigma'_v = 208.0 - 63.77 = 144.2 kPa$
Formula
$\sigma'_h = K_{0} \sigma'_v$
Substituting
$\sigma'_h = 0.45(144.2) = 64.91 kPa$
Formula
$\tau_f = \sigma' \tan \phi'$
Substituting
$\tau_f = 144.2 \tan 33^{\circ} = 93.67 kPa$

Answer:

\sigma_v = 208.0 kPa, u = 63.8 kPa, \sigma'_v = 144.2 kPa, \sigma'_h = 64.9 kPa, \tau_f = 93.7 kPa

Why the other options are there

τ_f = 135.1 kPa (total stress used)
σ′_h = 93.6 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Vertical Stress Profiles with Surcharge

Vertical Stress Profiles with Surcharge

Core formulas for this FE topic

Worked exam-style examples

Vertical Stress Profiles with Surcharge

Core formulas for this FE topic

σ−u=σ′\sigma - u = \sigma'σ−u=σ′

S=SurchargeVerticalWaterVerticalS = Surcharge Vertical Water VerticalS=SurchargeVerticalWaterVertical

Hγ1σ=S+(γ×H)H \gamma 1 \sigma = S + (\gamma \times H )Hγ1σ=S+(γ×H)

0σ′=S+(γ1×H1)0 \sigma' = S + (\gamma_{1} \times H_{1})0σ′=S+(γ1​×H1​)

H2 γ2 u = γw × H2Needs technical review

γ2=saturatedσ=S+[(γ1×H1)+σ′=S+[(γ1×H1)\gamma_{2} = saturated \sigma = S + [(\gamma_{1} \times H_{1}) + \sigma' = S + [(\gamma_{1} \times H_{1})γ2​=saturatedσ=S+[(γ1​×H1​)+σ′=S+[(γ1​×H1​)

unit weight (γ2 × H2)] + (γ2 × H2)] − (γw × H2)Needs technical review

Worked exam-style examples