Shear stress at failure

Example 1

Vertical stress profile with surcharge and the shear stress at failure — Shear stress at failure

A profile has 3.0 m of moist sand (γ = 18.0 kN/m³) over 3.0 m of saturated sand (γ_sat = 19.0 kN/m³) with the water table at the interface. A surcharge of 31 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′ = 30°.

Given

Layer 1:
$3.0 m, \gamma = 18.0 kN/m^{3}$
Layer 2:
$3.0 m, \gamma_sat = 19.0 kN/m^{3}$
$q = 31 kPa$
$K_{0} = 0.45, \phi' = 30^{\circ}$

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
$\sigma_v = 31 + 18.0(3.0) + 19.0(3.0) = 142.0 kPa$
Formula
$u = \gamma_w h_w$
Substituting
$u = 9.81(3.0) = 29.43 kPa$
Formula
$\sigma'_v = \sigma_v - u$
Substituting
$\sigma'_v = 142.0 - 29.43 = 112.6 kPa$
Formula
$\sigma'_h = K_{0} \sigma'_v$
Substituting
$\sigma'_h = 0.45(112.6) = 50.66 kPa$
Formula
$\tau_f = \sigma' \tan \phi'$
Substituting
$\tau_f = 112.6 \tan 30^{\circ} = 64.99 kPa$

Answer:

\sigma_v = 142.0 kPa, u = 29.4 kPa, \sigma'_v = 112.6 kPa, \sigma'_h = 50.7 kPa, \tau_f = 65.0 kPa

Why the other options are there

τ_f = 82.0 kPa (total stress used)
σ′_h = 63.9 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Shear stress at failure

Example 2

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

A profile has 4.5 m of moist sand (γ = 17.5 kN/m³) over 7.5 m of saturated sand (γ_sat = 19.5 kN/m³) with the water table at the interface. A surcharge of 49 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:
$4.5 m, \gamma = 17.5 kN/m^{3}$
Layer 2:
$7.5 m, \gamma_sat = 19.5 kN/m^{3}$
$q = 49 kPa$
$K_{0} = 0.40, \phi' = 36^{\circ}$

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
$\sigma_v = 49 + 17.5(4.5) + 19.5(7.5) = 274.0 kPa$
Formula
$u = \gamma_w h_w$
Substituting
$u = 9.81(7.5) = 73.58 kPa$
Formula
$\sigma'_v = \sigma_v - u$
Substituting
$\sigma'_v = 274.0 - 73.58 = 200.4 kPa$
Formula
$\sigma'_h = K_{0} \sigma'_v$
Substituting
$\sigma'_h = 0.40(200.4) = 80.17 kPa$
Formula
$\tau_f = \sigma' \tan \phi'$
Substituting
$\tau_f = 200.4 \tan 36^{\circ} = 145.6 kPa$

Answer:

\sigma_v = 274.0 kPa, u = 73.6 kPa, \sigma'_v = 200.4 kPa, \sigma'_h = 80.2 kPa, \tau_f = 145.6 kPa

Why the other options are there

τ_f = 199.1 kPa (total stress used)
σ′_h = 109.6 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Shear stress at failure

Example 3

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

A profile has 5.0 m of moist sand (γ = 18.5 kN/m³) over 7.5 m of saturated sand (γ_sat = 19.5 kN/m³) with the water table at the interface. A surcharge of 37 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′ = 36°.

Given

Layer 1:
$5.0 m, \gamma = 18.5 kN/m^{3}$
Layer 2:
$7.5 m, \gamma_sat = 19.5 kN/m^{3}$
$q = 37 kPa$
$K_{0} = 0.45, \phi' = 36^{\circ}$

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
$\sigma_v = 37 + 18.5(5.0) + 19.5(7.5) = 275.8 kPa$
Formula
$u = \gamma_w h_w$
Substituting
$u = 9.81(7.5) = 73.58 kPa$
Formula
$\sigma'_v = \sigma_v - u$
Substituting
$\sigma'_v = 275.8 - 73.58 = 202.2 kPa$
Formula
$\sigma'_h = K_{0} \sigma'_v$
Substituting
$\sigma'_h = 0.45(202.2) = 90.98 kPa$
Formula
$\tau_f = \sigma' \tan \phi'$
Substituting
$\tau_f = 202.2 \tan 36^{\circ} = 146.9 kPa$

Answer:

\sigma_v = 275.8 kPa, u = 73.6 kPa, \sigma'_v = 202.2 kPa, \sigma'_h = 91.0 kPa, \tau_f = 146.9 kPa

Why the other options are there

τ_f = 200.3 kPa (total stress used)
σ′_h = 124.1 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Shear stress at failure

Example 4

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

A profile has 5.0 m of moist sand (γ = 17.5 kN/m³) over 6.5 m of saturated sand (γ_sat = 19.5 kN/m³) with the water table at the interface. A surcharge of 22 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:
$5.0 m, \gamma = 17.5 kN/m^{3}$
Layer 2:
$6.5 m, \gamma_sat = 19.5 kN/m^{3}$
$q = 22 kPa$
$K_{0} = 0.60, \phi' = 31^{\circ}$

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
$\sigma_v = 22 + 17.5(5.0) + 19.5(6.5) = 236.3 kPa$
Formula
$u = \gamma_w h_w$
Substituting
$u = 9.81(6.5) = 63.77 kPa$
Formula
$\sigma'_v = \sigma_v - u$
Substituting
$\sigma'_v = 236.3 - 63.77 = 172.5 kPa$
Formula
$\sigma'_h = K_{0} \sigma'_v$
Substituting
$\sigma'_h = 0.60(172.5) = 103.5 kPa$
Formula
$\tau_f = \sigma' \tan \phi'$
Substituting
$\tau_f = 172.5 \tan 31^{\circ} = 103.6 kPa$

Answer:

\sigma_v = 236.3 kPa, u = 63.8 kPa, \sigma'_v = 172.5 kPa, \sigma'_h = 103.5 kPa, \tau_f = 103.6 kPa

Why the other options are there

τ_f = 142.0 kPa (total stress used)
σ′_h = 141.8 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Shear stress at failure

Example 5

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

A profile has 4.0 m of moist sand (γ = 17.5 kN/m³) over 4.5 m of saturated sand (γ_sat = 19.0 kN/m³) with the water table at the interface. A surcharge of 10 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′ = 36°.

Given

Layer 1:
$4.0 m, \gamma = 17.5 kN/m^{3}$
Layer 2:
$4.5 m, \gamma_sat = 19.0 kN/m^{3}$
$q = 10 kPa$
$K_{0} = 0.50, \phi' = 36^{\circ}$

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
$\sigma_v = 10 + 17.5(4.0) + 19.0(4.5) = 165.5 kPa$
Formula
$u = \gamma_w h_w$
Substituting
$u = 9.81(4.5) = 44.15 kPa$
Formula
$\sigma'_v = \sigma_v - u$
Substituting
$\sigma'_v = 165.5 - 44.15 = 121.4 kPa$
Formula
$\sigma'_h = K_{0} \sigma'_v$
Substituting
$\sigma'_h = 0.50(121.4) = 60.68 kPa$
Formula
$\tau_f = \sigma' \tan \phi'$
Substituting
$\tau_f = 121.4 \tan 36^{\circ} = 88.17 kPa$

Answer:

\sigma_v = 165.5 kPa, u = 44.1 kPa, \sigma'_v = 121.4 kPa, \sigma'_h = 60.7 kPa, \tau_f = 88.2 kPa

Why the other options are there

τ_f = 120.2 kPa (total stress used)
σ′_h = 82.8 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Shear stress at failure

Example 6

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

A profile has 6.0 m of moist sand (γ = 18.0 kN/m³) over 3.5 m of saturated sand (γ_sat = 21.5 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′ = 33°.

Given

Layer 1:
$6.0 m, \gamma = 18.0 kN/m^{3}$
Layer 2:
$3.5 m, \gamma_sat = 21.5 kN/m^{3}$
$q = 21 kPa$
$K_{0} = 0.45, \phi' = 33^{\circ}$

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
$\sigma_v = 21 + 18.0(6.0) + 21.5(3.5) = 204.3 kPa$
Formula
$u = \gamma_w h_w$
Substituting
$u = 9.81(3.5) = 34.34 kPa$
Formula
$\sigma'_v = \sigma_v - u$
Substituting
$\sigma'_v = 204.3 - 34.34 = 169.9 kPa$
Formula
$\sigma'_h = K_{0} \sigma'_v$
Substituting
$\sigma'_h = 0.45(169.9) = 76.46 kPa$
Formula
$\tau_f = \sigma' \tan \phi'$
Substituting
$\tau_f = 169.9 \tan 33^{\circ} = 110.3 kPa$

Answer:

\sigma_v = 204.3 kPa, u = 34.3 kPa, \sigma'_v = 169.9 kPa, \sigma'_h = 76.5 kPa, \tau_f = 110.3 kPa

Why the other options are there

τ_f = 132.6 kPa (total stress used)
σ′_h = 91.9 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Shear stress at failure

Example 7

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

A profile has 3.0 m of moist sand (γ = 20.0 kN/m³) over 3.5 m of saturated sand (γ_sat = 21.0 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 = 0.50) and the shear stress at failure for φ′ = 36°.

Given

Layer 1:
$3.0 m, \gamma = 20.0 kN/m^{3}$
Layer 2:
$3.5 m, \gamma_sat = 21.0 kN/m^{3}$
$q = 40 kPa$
$K_{0} = 0.50, \phi' = 36^{\circ}$

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
$\sigma_v = 40 + 20.0(3.0) + 21.0(3.5) = 173.5 kPa$
Formula
$u = \gamma_w h_w$
Substituting
$u = 9.81(3.5) = 34.34 kPa$
Formula
$\sigma'_v = \sigma_v - u$
Substituting
$\sigma'_v = 173.5 - 34.34 = 139.2 kPa$
Formula
$\sigma'_h = K_{0} \sigma'_v$
Substituting
$\sigma'_h = 0.50(139.2) = 69.58 kPa$
Formula
$\tau_f = \sigma' \tan \phi'$
Substituting
$\tau_f = 139.2 \tan 36^{\circ} = 101.1 kPa$

Answer:

\sigma_v = 173.5 kPa, u = 34.3 kPa, \sigma'_v = 139.2 kPa, \sigma'_h = 69.6 kPa, \tau_f = 101.1 kPa

Why the other options are there

τ_f = 126.1 kPa (total stress used)
σ′_h = 86.8 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Shear stress at failure

Example 8

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

A profile has 5.0 m of moist sand (γ = 19.0 kN/m³) over 3.5 m of saturated sand (γ_sat = 19.5 kN/m³) with the water table at the interface. A surcharge of 57 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′ = 29°.

Given

Layer 1:
$5.0 m, \gamma = 19.0 kN/m^{3}$
Layer 2:
$3.5 m, \gamma_sat = 19.5 kN/m^{3}$
$q = 57 kPa$
$K_{0} = 0.50, \phi' = 29^{\circ}$

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
$\sigma_v = 57 + 19.0(5.0) + 19.5(3.5) = 220.3 kPa$
Formula
$u = \gamma_w h_w$
Substituting
$u = 9.81(3.5) = 34.34 kPa$
Formula
$\sigma'_v = \sigma_v - u$
Substituting
$\sigma'_v = 220.3 - 34.34 = 185.9 kPa$
Formula
$\sigma'_h = K_{0} \sigma'_v$
Substituting
$\sigma'_h = 0.50(185.9) = 92.96 kPa$
Formula
$\tau_f = \sigma' \tan \phi'$
Substituting
$\tau_f = 185.9 \tan 29^{\circ} = 103.1 kPa$

Answer:

\sigma_v = 220.3 kPa, u = 34.3 kPa, \sigma'_v = 185.9 kPa, \sigma'_h = 93.0 kPa, \tau_f = 103.1 kPa

Why the other options are there

τ_f = 122.1 kPa (total stress used)
σ′_h = 110.1 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Shear stress at failure

Example 9

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

A profile has 2.5 m of moist sand (γ = 19.0 kN/m³) over 7.0 m of saturated sand (γ_sat = 20.5 kN/m³) with the water table at the interface. A surcharge of 59 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′ = 36°.

Given

Layer 1:
$2.5 m, \gamma = 19.0 kN/m^{3}$
Layer 2:
$7.0 m, \gamma_sat = 20.5 kN/m^{3}$
$q = 59 kPa$
$K_{0} = 0.45, \phi' = 36^{\circ}$

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
$\sigma_v = 59 + 19.0(2.5) + 20.5(7.0) = 250.0 kPa$
Formula
$u = \gamma_w h_w$
Substituting
$u = 9.81(7.0) = 68.67 kPa$
Formula
$\sigma'_v = \sigma_v - u$
Substituting
$\sigma'_v = 250.0 - 68.67 = 181.3 kPa$
Formula
$\sigma'_h = K_{0} \sigma'_v$
Substituting
$\sigma'_h = 0.45(181.3) = 81.60 kPa$
Formula
$\tau_f = \sigma' \tan \phi'$
Substituting
$\tau_f = 181.3 \tan 36^{\circ} = 131.7 kPa$

Answer:

\sigma_v = 250.0 kPa, u = 68.7 kPa, \sigma'_v = 181.3 kPa, \sigma'_h = 81.6 kPa, \tau_f = 131.7 kPa

Why the other options are there

τ_f = 181.6 kPa (total stress used)
σ′_h = 112.5 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Shear stress at failure

Example 10

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

A profile has 3.5 m of moist sand (γ = 18.5 kN/m³) over 7.5 m of saturated sand (γ_sat = 19.0 kN/m³) with the water table at the interface. A surcharge of 8 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:
$3.5 m, \gamma = 18.5 kN/m^{3}$
Layer 2:
$7.5 m, \gamma_sat = 19.0 kN/m^{3}$
$q = 8 kPa$
$K_{0} = 0.45, \phi' = 26^{\circ}$

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
$\sigma_v = 8 + 18.5(3.5) + 19.0(7.5) = 215.3 kPa$
Formula
$u = \gamma_w h_w$
Substituting
$u = 9.81(7.5) = 73.58 kPa$
Formula
$\sigma'_v = \sigma_v - u$
Substituting
$\sigma'_v = 215.3 - 73.58 = 141.7 kPa$
Formula
$\sigma'_h = K_{0} \sigma'_v$
Substituting
$\sigma'_h = 0.45(141.7) = 63.75 kPa$
Formula
$\tau_f = \sigma' \tan \phi'$
Substituting
$\tau_f = 141.7 \tan 26^{\circ} = 69.10 kPa$

Answer:

\sigma_v = 215.3 kPa, u = 73.6 kPa, \sigma'_v = 141.7 kPa, \sigma'_h = 63.8 kPa, \tau_f = 69.1 kPa

Why the other options are there

τ_f = 105.0 kPa (total stress used)
σ′_h = 96.9 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Shear stress at failure

Core formulas for this FE topic

Worked exam-style examples