Shear stress

Example 1

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

A profile has 5.5 m of moist sand (γ = 19.0 kN/m³) over 5.0 m of saturated sand (γ_sat = 21.0 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 = 0.45) and the shear stress at failure for φ′ = 35°.

Given

Layer 1:
$5.5 m, \gamma = 19.0 kN/m^{3}$
Layer 2:
$5.0 m, \gamma_sat = 21.0 kN/m^{3}$
$q = 46 kPa$
$K_{0} = 0.45, \phi' = 35^{\circ}$

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
$\sigma_v = 46 + 19.0(5.5) + 21.0(5.0) = 255.5 kPa$
Formula
$u = \gamma_w h_w$
Substituting
$u = 9.81(5.0) = 49.05 kPa$
Formula
$\sigma'_v = \sigma_v - u$
Substituting
$\sigma'_v = 255.5 - 49.05 = 206.5 kPa$
Formula
$\sigma'_h = K_{0} \sigma'_v$
Substituting
$\sigma'_h = 0.45(206.5) = 92.90 kPa$
Formula
$\tau_f = \sigma' \tan \phi'$
Substituting
$\tau_f = 206.5 \tan 35^{\circ} = 144.6 kPa$

Answer:

\sigma_v = 255.5 kPa, u = 49.1 kPa, \sigma'_v = 206.5 kPa, \sigma'_h = 92.9 kPa, \tau_f = 144.6 kPa

Why the other options are there

τ_f = 178.9 kPa (total stress used)
σ′_h = 115.0 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Shear stress

Example 2

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

A profile has 4.5 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 56 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′ = 33°.

Given

Layer 1:
$4.5 m, \gamma = 17.5 kN/m^{3}$
Layer 2:
$6.5 m, \gamma_sat = 19.5 kN/m^{3}$
$q = 56 kPa$
$K_{0} = 0.50, \phi' = 33^{\circ}$

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
$\sigma_v = 56 + 17.5(4.5) + 19.5(6.5) = 261.5 kPa$
Formula
$u = \gamma_w h_w$
Substituting
$u = 9.81(6.5) = 63.77 kPa$
Formula
$\sigma'_v = \sigma_v - u$
Substituting
$\sigma'_v = 261.5 - 63.77 = 197.7 kPa$
Formula
$\sigma'_h = K_{0} \sigma'_v$
Substituting
$\sigma'_h = 0.50(197.7) = 98.87 kPa$
Formula
$\tau_f = \sigma' \tan \phi'$
Substituting
$\tau_f = 197.7 \tan 33^{\circ} = 128.4 kPa$

Answer:

\sigma_v = 261.5 kPa, u = 63.8 kPa, \sigma'_v = 197.7 kPa, \sigma'_h = 98.9 kPa, \tau_f = 128.4 kPa

Why the other options are there

τ_f = 169.8 kPa (total stress used)
σ′_h = 130.8 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Shear stress

Example 3

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

A profile has 4.0 m of moist sand (γ = 19.5 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 14 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:
$4.0 m, \gamma = 19.5 kN/m^{3}$
Layer 2:
$7.5 m, \gamma_sat = 19.0 kN/m^{3}$
$q = 14 kPa$
$K_{0} = 0.45, \phi' = 27^{\circ}$

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
$\sigma_v = 14 + 19.5(4.0) + 19.0(7.5) = 234.5 kPa$
Formula
$u = \gamma_w h_w$
Substituting
$u = 9.81(7.5) = 73.58 kPa$
Formula
$\sigma'_v = \sigma_v - u$
Substituting
$\sigma'_v = 234.5 - 73.58 = 160.9 kPa$
Formula
$\sigma'_h = K_{0} \sigma'_v$
Substituting
$\sigma'_h = 0.45(160.9) = 72.42 kPa$
Formula
$\tau_f = \sigma' \tan \phi'$
Substituting
$\tau_f = 160.9 \tan 27^{\circ} = 82.00 kPa$

Answer:

\sigma_v = 234.5 kPa, u = 73.6 kPa, \sigma'_v = 160.9 kPa, \sigma'_h = 72.4 kPa, \tau_f = 82.0 kPa

Why the other options are there

τ_f = 119.5 kPa (total stress used)
σ′_h = 105.5 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Shear stress

Example 4

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

A profile has 5.5 m of moist sand (γ = 17.5 kN/m³) over 7.0 m of saturated sand (γ_sat = 18.5 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′ = 30°.

Given

Layer 1:
$5.5 m, \gamma = 17.5 kN/m^{3}$
Layer 2:
$7.0 m, \gamma_sat = 18.5 kN/m^{3}$
$q = 3 kPa$
$K_{0} = 0.55, \phi' = 30^{\circ}$

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
$\sigma_v = 3 + 17.5(5.5) + 18.5(7.0) = 228.8 kPa$
Formula
$u = \gamma_w h_w$
Substituting
$u = 9.81(7.0) = 68.67 kPa$
Formula
$\sigma'_v = \sigma_v - u$
Substituting
$\sigma'_v = 228.8 - 68.67 = 160.1 kPa$
Formula
$\sigma'_h = K_{0} \sigma'_v$
Substituting
$\sigma'_h = 0.55(160.1) = 88.04 kPa$
Formula
$\tau_f = \sigma' \tan \phi'$
Substituting
$\tau_f = 160.1 \tan 30^{\circ} = 92.42 kPa$

Answer:

\sigma_v = 228.8 kPa, u = 68.7 kPa, \sigma'_v = 160.1 kPa, \sigma'_h = 88.0 kPa, \tau_f = 92.4 kPa

Why the other options are there

τ_f = 132.1 kPa (total stress used)
σ′_h = 125.8 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Shear stress

Example 5

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

A profile has 2.5 m of moist sand (γ = 17.5 kN/m³) over 5.5 m of saturated sand (γ_sat = 19.5 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:
$2.5 m, \gamma = 17.5 kN/m^{3}$
Layer 2:
$5.5 m, \gamma_sat = 19.5 kN/m^{3}$
$q = 21 kPa$
$K_{0} = 0.45, \phi' = 33^{\circ}$

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
$\sigma_v = 21 + 17.5(2.5) + 19.5(5.5) = 172.0 kPa$
Formula
$u = \gamma_w h_w$
Substituting
$u = 9.81(5.5) = 53.96 kPa$
Formula
$\sigma'_v = \sigma_v - u$
Substituting
$\sigma'_v = 172.0 - 53.96 = 118.0 kPa$
Formula
$\sigma'_h = K_{0} \sigma'_v$
Substituting
$\sigma'_h = 0.45(118.0) = 53.12 kPa$
Formula
$\tau_f = \sigma' \tan \phi'$
Substituting
$\tau_f = 118.0 \tan 33^{\circ} = 76.66 kPa$

Answer:

\sigma_v = 172.0 kPa, u = 54.0 kPa, \sigma'_v = 118.0 kPa, \sigma'_h = 53.1 kPa, \tau_f = 76.7 kPa

Why the other options are there

τ_f = 111.7 kPa (total stress used)
σ′_h = 77.4 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Shear stress

Example 6

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

A profile has 4.5 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 12 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′ = 30°.

Given

Layer 1:
$4.5 m, \gamma = 19.0 kN/m^{3}$
Layer 2:
$3.5 m, \gamma_sat = 19.5 kN/m^{3}$
$q = 12 kPa$
$K_{0} = 0.50, \phi' = 30^{\circ}$

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
$\sigma_v = 12 + 19.0(4.5) + 19.5(3.5) = 165.8 kPa$
Formula
$u = \gamma_w h_w$
Substituting
$u = 9.81(3.5) = 34.34 kPa$
Formula
$\sigma'_v = \sigma_v - u$
Substituting
$\sigma'_v = 165.8 - 34.34 = 131.4 kPa$
Formula
$\sigma'_h = K_{0} \sigma'_v$
Substituting
$\sigma'_h = 0.50(131.4) = 65.71 kPa$
Formula
$\tau_f = \sigma' \tan \phi'$
Substituting
$\tau_f = 131.4 \tan 30^{\circ} = 75.87 kPa$

Answer:

\sigma_v = 165.8 kPa, u = 34.3 kPa, \sigma'_v = 131.4 kPa, \sigma'_h = 65.7 kPa, \tau_f = 75.9 kPa

Why the other options are there

τ_f = 95.7 kPa (total stress used)
σ′_h = 82.9 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Shear stress

Example 7

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

A profile has 5.5 m of moist sand (γ = 19.5 kN/m³) over 6.0 m of saturated sand (γ_sat = 21.5 kN/m³) with the water table at the interface. A surcharge of 14 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:
$5.5 m, \gamma = 19.5 kN/m^{3}$
Layer 2:
$6.0 m, \gamma_sat = 21.5 kN/m^{3}$
$q = 14 kPa$
$K_{0} = 0.40, \phi' = 33^{\circ}$

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
$\sigma_v = 14 + 19.5(5.5) + 21.5(6.0) = 250.3 kPa$
Formula
$u = \gamma_w h_w$
Substituting
$u = 9.81(6.0) = 58.86 kPa$
Formula
$\sigma'_v = \sigma_v - u$
Substituting
$\sigma'_v = 250.3 - 58.86 = 191.4 kPa$
Formula
$\sigma'_h = K_{0} \sigma'_v$
Substituting
$\sigma'_h = 0.40(191.4) = 76.56 kPa$
Formula
$\tau_f = \sigma' \tan \phi'$
Substituting
$\tau_f = 191.4 \tan 33^{\circ} = 124.3 kPa$

Answer:

\sigma_v = 250.3 kPa, u = 58.9 kPa, \sigma'_v = 191.4 kPa, \sigma'_h = 76.6 kPa, \tau_f = 124.3 kPa

Why the other options are there

τ_f = 162.5 kPa (total stress used)
σ′_h = 100.1 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Shear stress

Example 8

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

A profile has 3.0 m of moist sand (γ = 18.5 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 19 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′ = 33°.

Given

Layer 1:
$3.0 m, \gamma = 18.5 kN/m^{3}$
Layer 2:
$7.0 m, \gamma_sat = 21.0 kN/m^{3}$
$q = 19 kPa$
$K_{0} = 0.50, \phi' = 33^{\circ}$

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
$\sigma_v = 19 + 18.5(3.0) + 21.0(7.0) = 221.5 kPa$
Formula
$u = \gamma_w h_w$
Substituting
$u = 9.81(7.0) = 68.67 kPa$
Formula
$\sigma'_v = \sigma_v - u$
Substituting
$\sigma'_v = 221.5 - 68.67 = 152.8 kPa$
Formula
$\sigma'_h = K_{0} \sigma'_v$
Substituting
$\sigma'_h = 0.50(152.8) = 76.41 kPa$
Formula
$\tau_f = \sigma' \tan \phi'$
Substituting
$\tau_f = 152.8 \tan 33^{\circ} = 99.25 kPa$

Answer:

\sigma_v = 221.5 kPa, u = 68.7 kPa, \sigma'_v = 152.8 kPa, \sigma'_h = 76.4 kPa, \tau_f = 99.2 kPa

Why the other options are there

τ_f = 143.8 kPa (total stress used)
σ′_h = 110.8 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Shear stress

Example 9

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

A profile has 4.5 m of moist sand (γ = 17.0 kN/m³) over 8.5 m of saturated sand (γ_sat = 19.5 kN/m³) with the water table at the interface. A surcharge of 18 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:
$4.5 m, \gamma = 17.0 kN/m^{3}$
Layer 2:
$8.5 m, \gamma_sat = 19.5 kN/m^{3}$
$q = 18 kPa$
$K_{0} = 0.45, \phi' = 26^{\circ}$

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
$\sigma_v = 18 + 17.0(4.5) + 19.5(8.5) = 260.3 kPa$
Formula
$u = \gamma_w h_w$
Substituting
$u = 9.81(8.5) = 83.39 kPa$
Formula
$\sigma'_v = \sigma_v - u$
Substituting
$\sigma'_v = 260.3 - 83.39 = 176.9 kPa$
Formula
$\sigma'_h = K_{0} \sigma'_v$
Substituting
$\sigma'_h = 0.45(176.9) = 79.59 kPa$
Formula
$\tau_f = \sigma' \tan \phi'$
Substituting
$\tau_f = 176.9 \tan 26^{\circ} = 86.26 kPa$

Answer:

\sigma_v = 260.3 kPa, u = 83.4 kPa, \sigma'_v = 176.9 kPa, \sigma'_h = 79.6 kPa, \tau_f = 86.3 kPa

Why the other options are there

τ_f = 126.9 kPa (total stress used)
σ′_h = 117.1 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Shear stress

Example 10

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

A profile has 5.5 m of moist sand (γ = 18.0 kN/m³) over 6.5 m of saturated sand (γ_sat = 18.5 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 = 0.50) and the shear stress at failure for φ′ = 26°.

Given

Layer 1:
$5.5 m, \gamma = 18.0 kN/m^{3}$
Layer 2:
$6.5 m, \gamma_sat = 18.5 kN/m^{3}$
$q = 19 kPa$
$K_{0} = 0.50, \phi' = 26^{\circ}$

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
$\sigma_v = 19 + 18.0(5.5) + 18.5(6.5) = 238.3 kPa$
Formula
$u = \gamma_w h_w$
Substituting
$u = 9.81(6.5) = 63.77 kPa$
Formula
$\sigma'_v = \sigma_v - u$
Substituting
$\sigma'_v = 238.3 - 63.77 = 174.5 kPa$
Formula
$\sigma'_h = K_{0} \sigma'_v$
Substituting
$\sigma'_h = 0.50(174.5) = 87.24 kPa$
Formula
$\tau_f = \sigma' \tan \phi'$
Substituting
$\tau_f = 174.5 \tan 26^{\circ} = 85.10 kPa$

Answer:

\sigma_v = 238.3 kPa, u = 63.8 kPa, \sigma'_v = 174.5 kPa, \sigma'_h = 87.2 kPa, \tau_f = 85.1 kPa

Why the other options are there

τ_f = 116.2 kPa (total stress used)
σ′_h = 119.1 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Shear stress

Core formulas for this FE topic

Worked exam-style examples

Shear stress

Core formulas for this FE topic

τ=T/A\tau = T/Aτ=T/A

whereT=shearingforcewhere T = shearing forcewhereT=shearingforce

Worked exam-style examples