Vertical Stress Profiles

Example 1

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

A profile has 5.5 m of moist sand (γ = 20.0 kN/m³) over 5.0 m of saturated sand (γ_sat = 20.0 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′ = 32°.

Given

Layer 1:
$5.5 m, \gamma = 20.0 kN/m^{3}$
Layer 2:
$5.0 m, \gamma_sat = 20.0 kN/m^{3}$
$q = 19 kPa$
$K_{0} = 0.45, \phi' = 32^{\circ}$

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
$\sigma_v = 19 + 20.0(5.5) + 20.0(5.0) = 229.0 kPa$
Formula
$u = \gamma_w h_w$
Substituting
$u = 9.81(5.0) = 49.05 kPa$
Formula
$\sigma'_v = \sigma_v - u$
Substituting
$\sigma'_v = 229.0 - 49.05 = 180.0 kPa$
Formula
$\sigma'_h = K_{0} \sigma'_v$
Substituting
$\sigma'_h = 0.45(180.0) = 80.98 kPa$
Formula
$\tau_f = \sigma' \tan \phi'$
Substituting
$\tau_f = 180.0 \tan 32^{\circ} = 112.4 kPa$

Answer:

\sigma_v = 229.0 kPa, u = 49.1 kPa, \sigma'_v = 180.0 kPa, \sigma'_h = 81.0 kPa, \tau_f = 112.4 kPa

Why the other options are there

τ_f = 143.1 kPa (total stress used)
σ′_h = 103.1 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Vertical Stress Profiles

Example 2

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

A profile has 4.5 m of moist sand (γ = 19.0 kN/m³) over 4.0 m of saturated sand (γ_sat = 21.0 kN/m³) with the water table at the interface. A surcharge of 52 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′ = 28°.

Given

Layer 1:
$4.5 m, \gamma = 19.0 kN/m^{3}$
Layer 2:
$4.0 m, \gamma_sat = 21.0 kN/m^{3}$
$q = 52 kPa$
$K_{0} = 0.45, \phi' = 28^{\circ}$

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
$\sigma_v = 52 + 19.0(4.5) + 21.0(4.0) = 221.5 kPa$
Formula
$u = \gamma_w h_w$
Substituting
$u = 9.81(4.0) = 39.24 kPa$
Formula
$\sigma'_v = \sigma_v - u$
Substituting
$\sigma'_v = 221.5 - 39.24 = 182.3 kPa$
Formula
$\sigma'_h = K_{0} \sigma'_v$
Substituting
$\sigma'_h = 0.45(182.3) = 82.02 kPa$
Formula
$\tau_f = \sigma' \tan \phi'$
Substituting
$\tau_f = 182.3 \tan 28^{\circ} = 96.91 kPa$

Answer:

\sigma_v = 221.5 kPa, u = 39.2 kPa, \sigma'_v = 182.3 kPa, \sigma'_h = 82.0 kPa, \tau_f = 96.9 kPa

Why the other options are there

τ_f = 117.8 kPa (total stress used)
σ′_h = 99.7 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Vertical Stress Profiles

Example 3

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

A profile has 2.5 m of moist sand (γ = 17.0 kN/m³) over 8.0 m of saturated sand (γ_sat = 18.5 kN/m³) with the water table at the interface. A surcharge of 48 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:
$2.5 m, \gamma = 17.0 kN/m^{3}$
Layer 2:
$8.0 m, \gamma_sat = 18.5 kN/m^{3}$
$q = 48 kPa$
$K_{0} = 0.60, \phi' = 30^{\circ}$

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
$\sigma_v = 48 + 17.0(2.5) + 18.5(8.0) = 238.5 kPa$
Formula
$u = \gamma_w h_w$
Substituting
$u = 9.81(8.0) = 78.48 kPa$
Formula
$\sigma'_v = \sigma_v - u$
Substituting
$\sigma'_v = 238.5 - 78.48 = 160.0 kPa$
Formula
$\sigma'_h = K_{0} \sigma'_v$
Substituting
$\sigma'_h = 0.60(160.0) = 96.01 kPa$
Formula
$\tau_f = \sigma' \tan \phi'$
Substituting
$\tau_f = 160.0 \tan 30^{\circ} = 92.39 kPa$

Answer:

\sigma_v = 238.5 kPa, u = 78.5 kPa, \sigma'_v = 160.0 kPa, \sigma'_h = 96.0 kPa, \tau_f = 92.4 kPa

Why the other options are there

τ_f = 137.7 kPa (total stress used)
σ′_h = 143.1 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Vertical Stress Profiles

Example 4

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

A profile has 3.5 m of moist sand (γ = 18.0 kN/m³) over 6.5 m of saturated sand (γ_sat = 19.0 kN/m³) with the water table at the interface. A surcharge of 27 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′ = 26°.

Given

Layer 1:
$3.5 m, \gamma = 18.0 kN/m^{3}$
Layer 2:
$6.5 m, \gamma_sat = 19.0 kN/m^{3}$
$q = 27 kPa$
$K_{0} = 0.45, \phi' = 26^{\circ}$

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
$\sigma_v = 27 + 18.0(3.5) + 19.0(6.5) = 213.5 kPa$
Formula
$u = \gamma_w h_w$
Substituting
$u = 9.81(6.5) = 63.77 kPa$
Formula
$\sigma'_v = \sigma_v - u$
Substituting
$\sigma'_v = 213.5 - 63.77 = 149.7 kPa$
Formula
$\sigma'_h = K_{0} \sigma'_v$
Substituting
$\sigma'_h = 0.45(149.7) = 67.38 kPa$
Formula
$\tau_f = \sigma' \tan \phi'$
Substituting
$\tau_f = 149.7 \tan 26^{\circ} = 73.03 kPa$

Answer:

\sigma_v = 213.5 kPa, u = 63.8 kPa, \sigma'_v = 149.7 kPa, \sigma'_h = 67.4 kPa, \tau_f = 73.0 kPa

Why the other options are there

τ_f = 104.1 kPa (total stress used)
σ′_h = 96.1 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Vertical Stress Profiles

Example 5

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

A profile has 6.0 m of moist sand (γ = 18.5 kN/m³) over 5.0 m of saturated sand (γ_sat = 20.0 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 = 0.40) and the shear stress at failure for φ′ = 30°.

Given

Layer 1:
$6.0 m, \gamma = 18.5 kN/m^{3}$
Layer 2:
$5.0 m, \gamma_sat = 20.0 kN/m^{3}$
$q = 25 kPa$
$K_{0} = 0.40, \phi' = 30^{\circ}$

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
$\sigma_v = 25 + 18.5(6.0) + 20.0(5.0) = 236.0 kPa$
Formula
$u = \gamma_w h_w$
Substituting
$u = 9.81(5.0) = 49.05 kPa$
Formula
$\sigma'_v = \sigma_v - u$
Substituting
$\sigma'_v = 236.0 - 49.05 = 187.0 kPa$
Formula
$\sigma'_h = K_{0} \sigma'_v$
Substituting
$\sigma'_h = 0.40(187.0) = 74.78 kPa$
Formula
$\tau_f = \sigma' \tan \phi'$
Substituting
$\tau_f = 187.0 \tan 30^{\circ} = 107.9 kPa$

Answer:

\sigma_v = 236.0 kPa, u = 49.1 kPa, \sigma'_v = 187.0 kPa, \sigma'_h = 74.8 kPa, \tau_f = 107.9 kPa

Why the other options are there

τ_f = 136.3 kPa (total stress used)
σ′_h = 94.4 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Vertical Stress Profiles

Example 6

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

A profile has 2.0 m of moist sand (γ = 19.0 kN/m³) over 7.5 m of saturated sand (γ_sat = 20.5 kN/m³) with the water table at the interface. A surcharge of 41 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′ = 34°.

Given

Layer 1:
$2.0 m, \gamma = 19.0 kN/m^{3}$
Layer 2:
$7.5 m, \gamma_sat = 20.5 kN/m^{3}$
$q = 41 kPa$
$K_{0} = 0.45, \phi' = 34^{\circ}$

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
$\sigma_v = 41 + 19.0(2.0) + 20.5(7.5) = 232.8 kPa$
Formula
$u = \gamma_w h_w$
Substituting
$u = 9.81(7.5) = 73.58 kPa$
Formula
$\sigma'_v = \sigma_v - u$
Substituting
$\sigma'_v = 232.8 - 73.58 = 159.2 kPa$
Formula
$\sigma'_h = K_{0} \sigma'_v$
Substituting
$\sigma'_h = 0.45(159.2) = 71.63 kPa$
Formula
$\tau_f = \sigma' \tan \phi'$
Substituting
$\tau_f = 159.2 \tan 34^{\circ} = 107.4 kPa$

Answer:

\sigma_v = 232.8 kPa, u = 73.6 kPa, \sigma'_v = 159.2 kPa, \sigma'_h = 71.6 kPa, \tau_f = 107.4 kPa

Why the other options are there

τ_f = 157.0 kPa (total stress used)
σ′_h = 104.7 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Vertical Stress Profiles

Example 7

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

A profile has 4.0 m of moist sand (γ = 19.0 kN/m³) over 7.5 m of saturated sand (γ_sat = 21.0 kN/m³) with the water table at the interface. A surcharge of 38 kPa acts at the surface. At the base compute the total normal stress, pore pressure, effective stress, horizontal stress (K₀ = 0.50) and the shear stress at failure for φ′ = 32°.

Given

Layer 1:
$4.0 m, \gamma = 19.0 kN/m^{3}$
Layer 2:
$7.5 m, \gamma_sat = 21.0 kN/m^{3}$
$q = 38 kPa$
$K_{0} = 0.50, \phi' = 32^{\circ}$

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
$\sigma_v = 38 + 19.0(4.0) + 21.0(7.5) = 271.5 kPa$
Formula
$u = \gamma_w h_w$
Substituting
$u = 9.81(7.5) = 73.58 kPa$
Formula
$\sigma'_v = \sigma_v - u$
Substituting
$\sigma'_v = 271.5 - 73.58 = 197.9 kPa$
Formula
$\sigma'_h = K_{0} \sigma'_v$
Substituting
$\sigma'_h = 0.50(197.9) = 98.96 kPa$
Formula
$\tau_f = \sigma' \tan \phi'$
Substituting
$\tau_f = 197.9 \tan 32^{\circ} = 123.7 kPa$

Answer:

\sigma_v = 271.5 kPa, u = 73.6 kPa, \sigma'_v = 197.9 kPa, \sigma'_h = 99.0 kPa, \tau_f = 123.7 kPa

Why the other options are there

τ_f = 169.7 kPa (total stress used)
σ′_h = 135.8 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Vertical Stress Profiles

Example 8

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

A profile has 5.0 m of moist sand (γ = 19.5 kN/m³) over 4.0 m of saturated sand (γ_sat = 21.0 kN/m³) with the water table at the interface. A surcharge of 48 kPa acts at the surface. At the base compute the total normal stress, pore pressure, effective stress, horizontal stress (K₀ = 0.50) and the shear stress at failure for φ′ = 35°.

Given

Layer 1:
$5.0 m, \gamma = 19.5 kN/m^{3}$
Layer 2:
$4.0 m, \gamma_sat = 21.0 kN/m^{3}$
$q = 48 kPa$
$K_{0} = 0.50, \phi' = 35^{\circ}$

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
$\sigma_v = 48 + 19.5(5.0) + 21.0(4.0) = 229.5 kPa$
Formula
$u = \gamma_w h_w$
Substituting
$u = 9.81(4.0) = 39.24 kPa$
Formula
$\sigma'_v = \sigma_v - u$
Substituting
$\sigma'_v = 229.5 - 39.24 = 190.3 kPa$
Formula
$\sigma'_h = K_{0} \sigma'_v$
Substituting
$\sigma'_h = 0.50(190.3) = 95.13 kPa$
Formula
$\tau_f = \sigma' \tan \phi'$
Substituting
$\tau_f = 190.3 \tan 35^{\circ} = 133.2 kPa$

Answer:

\sigma_v = 229.5 kPa, u = 39.2 kPa, \sigma'_v = 190.3 kPa, \sigma'_h = 95.1 kPa, \tau_f = 133.2 kPa

Why the other options are there

τ_f = 160.7 kPa (total stress used)
σ′_h = 114.8 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Vertical Stress Profiles

Example 9

Vertical stress profile with surcharge and the shear stress at failure — Vertical Stress Profiles (9)

A profile has 4.0 m of moist sand (γ = 18.0 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 36 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′ = 34°.

Given

Layer 1:
$4.0 m, \gamma = 18.0 kN/m^{3}$
Layer 2:
$7.5 m, \gamma_sat = 21.5 kN/m^{3}$
$q = 36 kPa$
$K_{0} = 0.40, \phi' = 34^{\circ}$

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
$\sigma_v = 36 + 18.0(4.0) + 21.5(7.5) = 269.3 kPa$
Formula
$u = \gamma_w h_w$
Substituting
$u = 9.81(7.5) = 73.58 kPa$
Formula
$\sigma'_v = \sigma_v - u$
Substituting
$\sigma'_v = 269.3 - 73.58 = 195.7 kPa$
Formula
$\sigma'_h = K_{0} \sigma'_v$
Substituting
$\sigma'_h = 0.40(195.7) = 78.27 kPa$
Formula
$\tau_f = \sigma' \tan \phi'$
Substituting
$\tau_f = 195.7 \tan 34^{\circ} = 132.0 kPa$

Answer:

\sigma_v = 269.3 kPa, u = 73.6 kPa, \sigma'_v = 195.7 kPa, \sigma'_h = 78.3 kPa, \tau_f = 132.0 kPa

Why the other options are there

τ_f = 181.6 kPa (total stress used)
σ′_h = 107.7 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Vertical Stress Profiles

Example 10

Vertical stress profile with surcharge and the shear stress at failure — Vertical Stress Profiles (10)

A profile has 3.0 m of moist sand (γ = 18.0 kN/m³) over 4.5 m of saturated sand (γ_sat = 21.0 kN/m³) with the water table at the interface. A surcharge of 54 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′ = 36°.

Given

Layer 1:
$3.0 m, \gamma = 18.0 kN/m^{3}$
Layer 2:
$4.5 m, \gamma_sat = 21.0 kN/m^{3}$
$q = 54 kPa$
$K_{0} = 0.55, \phi' = 36^{\circ}$

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
$\sigma_v = 54 + 18.0(3.0) + 21.0(4.5) = 202.5 kPa$
Formula
$u = \gamma_w h_w$
Substituting
$u = 9.81(4.5) = 44.15 kPa$
Formula
$\sigma'_v = \sigma_v - u$
Substituting
$\sigma'_v = 202.5 - 44.15 = 158.4 kPa$
Formula
$\sigma'_h = K_{0} \sigma'_v$
Substituting
$\sigma'_h = 0.55(158.4) = 87.10 kPa$
Formula
$\tau_f = \sigma' \tan \phi'$
Substituting
$\tau_f = 158.4 \tan 36^{\circ} = 115.1 kPa$

Answer:

\sigma_v = 202.5 kPa, u = 44.1 kPa, \sigma'_v = 158.4 kPa, \sigma'_h = 87.1 kPa, \tau_f = 115.1 kPa

Why the other options are there

τ_f = 147.1 kPa (total stress used)
σ′_h = 111.4 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Vertical Stress Profiles

Core formulas for this FE topic

Worked exam-style examples