Total normal stress

Example 1

Vertical stress profile with surcharge and the shear stress at failure — Total normal stress

A profile has 4.5 m of moist sand (γ = 19.5 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 9 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′ = 33°.

Given

Layer 1:
$4.5 m, \gamma = 19.5 kN/m^{3}$
Layer 2:
$6.5 m, \gamma_sat = 19.0 kN/m^{3}$
$q = 9 kPa$
$K_{0} = 0.55, \phi' = 33^{\circ}$

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
$\sigma_v = 9 + 19.5(4.5) + 19.0(6.5) = 220.3 kPa$
Formula
$u = \gamma_w h_w$
Substituting
$u = 9.81(6.5) = 63.77 kPa$
Formula
$\sigma'_v = \sigma_v - u$
Substituting
$\sigma'_v = 220.3 - 63.77 = 156.5 kPa$
Formula
$\sigma'_h = K_{0} \sigma'_v$
Substituting
$\sigma'_h = 0.55(156.5) = 86.07 kPa$
Formula
$\tau_f = \sigma' \tan \phi'$
Substituting
$\tau_f = 156.5 \tan 33^{\circ} = 101.6 kPa$

Answer:

\sigma_v = 220.3 kPa, u = 63.8 kPa, \sigma'_v = 156.5 kPa, \sigma'_h = 86.1 kPa, \tau_f = 101.6 kPa

Why the other options are there

τ_f = 143.0 kPa (total stress used)
σ′_h = 121.1 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Total normal stress

Example 2

Vertical stress profile with surcharge and the shear stress at failure — Total normal stress (2)

A profile has 5.5 m of moist sand (γ = 17.5 kN/m³) over 7.0 m of saturated sand (γ_sat = 21.0 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 = 0.40) and the shear stress at failure for φ′ = 33°.

Given

Layer 1:
$5.5 m, \gamma = 17.5 kN/m^{3}$
Layer 2:
$7.0 m, \gamma_sat = 21.0 kN/m^{3}$
$q = 50 kPa$
$K_{0} = 0.40, \phi' = 33^{\circ}$

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
$\sigma_v = 50 + 17.5(5.5) + 21.0(7.0) = 293.3 kPa$
Formula
$u = \gamma_w h_w$
Substituting
$u = 9.81(7.0) = 68.67 kPa$
Formula
$\sigma'_v = \sigma_v - u$
Substituting
$\sigma'_v = 293.3 - 68.67 = 224.6 kPa$
Formula
$\sigma'_h = K_{0} \sigma'_v$
Substituting
$\sigma'_h = 0.40(224.6) = 89.83 kPa$
Formula
$\tau_f = \sigma' \tan \phi'$
Substituting
$\tau_f = 224.6 \tan 33^{\circ} = 145.8 kPa$

Answer:

\sigma_v = 293.3 kPa, u = 68.7 kPa, \sigma'_v = 224.6 kPa, \sigma'_h = 89.8 kPa, \tau_f = 145.8 kPa

Why the other options are there

τ_f = 190.4 kPa (total stress used)
σ′_h = 117.3 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Total normal stress

Example 3

Vertical stress profile with surcharge and the shear stress at failure — Total normal stress (3)

A profile has 2.5 m of moist sand (γ = 20.0 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 5 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′ = 27°.

Given

Layer 1:
$2.5 m, \gamma = 20.0 kN/m^{3}$
Layer 2:
$6.5 m, \gamma_sat = 19.0 kN/m^{3}$
$q = 5 kPa$
$K_{0} = 0.50, \phi' = 27^{\circ}$

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
$\sigma_v = 5 + 20.0(2.5) + 19.0(6.5) = 178.5 kPa$
Formula
$u = \gamma_w h_w$
Substituting
$u = 9.81(6.5) = 63.77 kPa$
Formula
$\sigma'_v = \sigma_v - u$
Substituting
$\sigma'_v = 178.5 - 63.77 = 114.7 kPa$
Formula
$\sigma'_h = K_{0} \sigma'_v$
Substituting
$\sigma'_h = 0.50(114.7) = 57.37 kPa$
Formula
$\tau_f = \sigma' \tan \phi'$
Substituting
$\tau_f = 114.7 \tan 27^{\circ} = 58.46 kPa$

Answer:

\sigma_v = 178.5 kPa, u = 63.8 kPa, \sigma'_v = 114.7 kPa, \sigma'_h = 57.4 kPa, \tau_f = 58.5 kPa

Why the other options are there

τ_f = 91.0 kPa (total stress used)
σ′_h = 89.3 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Total normal stress

Example 4

Vertical stress profile with surcharge and the shear stress at failure — Total normal stress (4)

A profile has 3.5 m of moist sand (γ = 17.0 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 11 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′ = 31°.

Given

Layer 1:
$3.5 m, \gamma = 17.0 kN/m^{3}$
Layer 2:
$6.0 m, \gamma_sat = 19.0 kN/m^{3}$
$q = 11 kPa$
$K_{0} = 0.55, \phi' = 31^{\circ}$

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
$\sigma_v = 11 + 17.0(3.5) + 19.0(6.0) = 184.5 kPa$
Formula
$u = \gamma_w h_w$
Substituting
$u = 9.81(6.0) = 58.86 kPa$
Formula
$\sigma'_v = \sigma_v - u$
Substituting
$\sigma'_v = 184.5 - 58.86 = 125.6 kPa$
Formula
$\sigma'_h = K_{0} \sigma'_v$
Substituting
$\sigma'_h = 0.55(125.6) = 69.10 kPa$
Formula
$\tau_f = \sigma' \tan \phi'$
Substituting
$\tau_f = 125.6 \tan 31^{\circ} = 75.49 kPa$

Answer:

\sigma_v = 184.5 kPa, u = 58.9 kPa, \sigma'_v = 125.6 kPa, \sigma'_h = 69.1 kPa, \tau_f = 75.5 kPa

Why the other options are there

τ_f = 110.9 kPa (total stress used)
σ′_h = 101.5 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Total normal stress

Example 5

Vertical stress profile with surcharge and the shear stress at failure — Total normal stress (5)

A profile has 3.0 m of moist sand (γ = 19.0 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 21 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′ = 29°.

Given

Layer 1:
$3.0 m, \gamma = 19.0 kN/m^{3}$
Layer 2:
$6.0 m, \gamma_sat = 19.0 kN/m^{3}$
$q = 21 kPa$
$K_{0} = 0.45, \phi' = 29^{\circ}$

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
$\sigma_v = 21 + 19.0(3.0) + 19.0(6.0) = 192.0 kPa$
Formula
$u = \gamma_w h_w$
Substituting
$u = 9.81(6.0) = 58.86 kPa$
Formula
$\sigma'_v = \sigma_v - u$
Substituting
$\sigma'_v = 192.0 - 58.86 = 133.1 kPa$
Formula
$\sigma'_h = K_{0} \sigma'_v$
Substituting
$\sigma'_h = 0.45(133.1) = 59.91 kPa$
Formula
$\tau_f = \sigma' \tan \phi'$
Substituting
$\tau_f = 133.1 \tan 29^{\circ} = 73.80 kPa$

Answer:

\sigma_v = 192.0 kPa, u = 58.9 kPa, \sigma'_v = 133.1 kPa, \sigma'_h = 59.9 kPa, \tau_f = 73.8 kPa

Why the other options are there

τ_f = 106.4 kPa (total stress used)
σ′_h = 86.4 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Total normal stress

Example 6

Vertical stress profile with surcharge and the shear stress at failure — Total normal stress (6)

A profile has 6.0 m of moist sand (γ = 17.5 kN/m³) over 4.0 m of saturated sand (γ_sat = 20.5 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 = 0.60) and the shear stress at failure for φ′ = 31°.

Given

Layer 1:
$6.0 m, \gamma = 17.5 kN/m^{3}$
Layer 2:
$4.0 m, \gamma_sat = 20.5 kN/m^{3}$
$q = 50 kPa$
$K_{0} = 0.60, \phi' = 31^{\circ}$

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
$\sigma_v = 50 + 17.5(6.0) + 20.5(4.0) = 237.0 kPa$
Formula
$u = \gamma_w h_w$
Substituting
$u = 9.81(4.0) = 39.24 kPa$
Formula
$\sigma'_v = \sigma_v - u$
Substituting
$\sigma'_v = 237.0 - 39.24 = 197.8 kPa$
Formula
$\sigma'_h = K_{0} \sigma'_v$
Substituting
$\sigma'_h = 0.60(197.8) = 118.7 kPa$
Formula
$\tau_f = \sigma' \tan \phi'$
Substituting
$\tau_f = 197.8 \tan 31^{\circ} = 118.8 kPa$

Answer:

\sigma_v = 237.0 kPa, u = 39.2 kPa, \sigma'_v = 197.8 kPa, \sigma'_h = 118.7 kPa, \tau_f = 118.8 kPa

Why the other options are there

τ_f = 142.4 kPa (total stress used)
σ′_h = 142.2 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Total normal stress

Example 7

Vertical stress profile with surcharge and the shear stress at failure — Total normal stress (7)

A profile has 5.0 m of moist sand (γ = 17.5 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 51 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′ = 30°.

Given

Layer 1:
$5.0 m, \gamma = 17.5 kN/m^{3}$
Layer 2:
$4.0 m, \gamma_sat = 21.0 kN/m^{3}$
$q = 51 kPa$
$K_{0} = 0.55, \phi' = 30^{\circ}$

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
$\sigma_v = 51 + 17.5(5.0) + 21.0(4.0) = 222.5 kPa$
Formula
$u = \gamma_w h_w$
Substituting
$u = 9.81(4.0) = 39.24 kPa$
Formula
$\sigma'_v = \sigma_v - u$
Substituting
$\sigma'_v = 222.5 - 39.24 = 183.3 kPa$
Formula
$\sigma'_h = K_{0} \sigma'_v$
Substituting
$\sigma'_h = 0.55(183.3) = 100.8 kPa$
Formula
$\tau_f = \sigma' \tan \phi'$
Substituting
$\tau_f = 183.3 \tan 30^{\circ} = 105.8 kPa$

Answer:

\sigma_v = 222.5 kPa, u = 39.2 kPa, \sigma'_v = 183.3 kPa, \sigma'_h = 100.8 kPa, \tau_f = 105.8 kPa

Why the other options are there

τ_f = 128.5 kPa (total stress used)
σ′_h = 122.4 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Total normal stress

Example 8

Vertical stress profile with surcharge and the shear stress at failure — Total normal stress (8)

A profile has 5.0 m of moist sand (γ = 18.0 kN/m³) over 4.0 m of saturated sand (γ_sat = 20.0 kN/m³) with the water table at the interface. A surcharge of 45 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:
$5.0 m, \gamma = 18.0 kN/m^{3}$
Layer 2:
$4.0 m, \gamma_sat = 20.0 kN/m^{3}$
$q = 45 kPa$
$K_{0} = 0.45, \phi' = 35^{\circ}$

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
$\sigma_v = 45 + 18.0(5.0) + 20.0(4.0) = 215.0 kPa$
Formula
$u = \gamma_w h_w$
Substituting
$u = 9.81(4.0) = 39.24 kPa$
Formula
$\sigma'_v = \sigma_v - u$
Substituting
$\sigma'_v = 215.0 - 39.24 = 175.8 kPa$
Formula
$\sigma'_h = K_{0} \sigma'_v$
Substituting
$\sigma'_h = 0.45(175.8) = 79.09 kPa$
Formula
$\tau_f = \sigma' \tan \phi'$
Substituting
$\tau_f = 175.8 \tan 35^{\circ} = 123.1 kPa$

Answer:

\sigma_v = 215.0 kPa, u = 39.2 kPa, \sigma'_v = 175.8 kPa, \sigma'_h = 79.1 kPa, \tau_f = 123.1 kPa

Why the other options are there

τ_f = 150.5 kPa (total stress used)
σ′_h = 96.8 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Total normal stress

Example 9

Vertical stress profile with surcharge and the shear stress at failure — Total normal stress (9)

A profile has 3.0 m of moist sand (γ = 17.5 kN/m³) over 4.5 m of saturated sand (γ_sat = 19.5 kN/m³) with the water table at the interface. A surcharge of 3 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′ = 31°.

Given

Layer 1:
$3.0 m, \gamma = 17.5 kN/m^{3}$
Layer 2:
$4.5 m, \gamma_sat = 19.5 kN/m^{3}$
$q = 3 kPa$
$K_{0} = 0.55, \phi' = 31^{\circ}$

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
$\sigma_v = 3 + 17.5(3.0) + 19.5(4.5) = 143.3 kPa$
Formula
$u = \gamma_w h_w$
Substituting
$u = 9.81(4.5) = 44.15 kPa$
Formula
$\sigma'_v = \sigma_v - u$
Substituting
$\sigma'_v = 143.3 - 44.15 = 99.10 kPa$
Formula
$\sigma'_h = K_{0} \sigma'_v$
Substituting
$\sigma'_h = 0.55(99.10) = 54.51 kPa$
Formula
$\tau_f = \sigma' \tan \phi'$
Substituting
$\tau_f = 99.10 \tan 31^{\circ} = 59.55 kPa$

Answer:

\sigma_v = 143.3 kPa, u = 44.1 kPa, \sigma'_v = 99.1 kPa, \sigma'_h = 54.5 kPa, \tau_f = 59.5 kPa

Why the other options are there

τ_f = 86.1 kPa (total stress used)
σ′_h = 78.8 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Total normal stress

Example 10

Vertical stress profile with surcharge and the shear stress at failure — Total normal stress (10)

A profile has 5.5 m of moist sand (γ = 17.0 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 60 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′ = 36°.

Given

Layer 1:
$5.5 m, \gamma = 17.0 kN/m^{3}$
Layer 2:
$6.5 m, \gamma_sat = 21.0 kN/m^{3}$
$q = 60 kPa$
$K_{0} = 0.60, \phi' = 36^{\circ}$

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
$\sigma_v = 60 + 17.0(5.5) + 21.0(6.5) = 290.0 kPa$
Formula
$u = \gamma_w h_w$
Substituting
$u = 9.81(6.5) = 63.77 kPa$
Formula
$\sigma'_v = \sigma_v - u$
Substituting
$\sigma'_v = 290.0 - 63.77 = 226.2 kPa$
Formula
$\sigma'_h = K_{0} \sigma'_v$
Substituting
$\sigma'_h = 0.60(226.2) = 135.7 kPa$
Formula
$\tau_f = \sigma' \tan \phi'$
Substituting
$\tau_f = 226.2 \tan 36^{\circ} = 164.4 kPa$

Answer:

\sigma_v = 290.0 kPa, u = 63.8 kPa, \sigma'_v = 226.2 kPa, \sigma'_h = 135.7 kPa, \tau_f = 164.4 kPa

Why the other options are there

τ_f = 210.7 kPa (total stress used)
σ′_h = 174.0 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Total normal stress

Core formulas for this FE topic

Worked exam-style examples