10.3 Bearing capacity, stress and slope stability

Achievable FE Civil

10. Soil mechanics

Our FE Civil course is now in "early access" - get 50% off for a limited time.

Bearing capacity, stress and slope stability

8 min read

Font

Discuss

Feedback

This chapter covers the following topics:

Bearing capacity
Vertical stress profiles
Vertical stress profiles with surcharge
Horizontal stress profiles and forces
Retaining walls
Slope stability

Bearing capacity

Definitions

Bearing capacity: Bearing capacity is the maximum pressure that soil can safely support from a foundation without experiencing shear failure or excessive settlement.

A common expression for the ultimate bearing capacity of a strip footing is:

$q_{u lt} = c N_{c} + γ^{'} D_{f} N_{q} + \frac{1}{2} γ^{'} B N_{γ}$

Where:

$c$ = cohesion of the soil
$γ^{'}$ = effective unit weight of the soil
$N_{c}$ , $N_{q}$ , $N_{γ}$ = bearing capacity factors, which depend on $ϕ$ . The FE Reference Handbook defines them but does not list their values, so a problem supplies them. The expression $N_{q} = e^{π t a n ϕ} tan^{2} (45° + \frac{ϕ}{2})$ (the form used in Meyerhof’s and Vesic’s methods, not Terzaghi’s) shows how $N_{q}$ follows from $ϕ$ .
$D_{f}$ = depth of footing below ground surface
$B$ = width of strip footing

Pitfall: below the water table, use effective stress and effective unit weight $γ^{'}$ , not total values - pore water pressure adds no shear strength, so using total stress overstates bearing capacity. Use a consistent unit system (SI or US customary) throughout a calculation. The Handbook gives the $q_{u lt}$ equation but no table of $N_{c}$ , $N_{q}$ , and $N_{γ}$ , so use the values the problem provides for the given $ϕ$ .

Example: ultimate bearing capacity

A strip footing has $B = 2$ m, $D_{f} = 1.5$ m, $c = 20$ kPa, and $γ^{'} = 18$ kN/m³. For $ϕ = 30°$ , Meyerhof’s bearing capacity factors are $N_{c} = 30.14$ , $N_{q} = 18.4$ , $N_{γ} = 15.67$ .

$q_{u lt} = (20) (30.14) + (18) (1.5) (18.4) + \frac{1}{2} (18) (2) (15.67) = 1, 381.7 kPa$

Answer: $q_{u lt} \approx 1, 381.7$ kPa

Vertical stress profiles

Definitions

Vertical stress: Vertical stress is the stress acting perpendicular to a horizontal plane in soil due to self-weight and applied loads.

Vertical stress generally increases with depth because more soil (and any applied loads) lies above the point of interest.

Vertical stress profiles with surcharge

Definitions

Surcharge: Surcharge is an additional surface load applied to the ground, such as traffic load or structural loading, that increases vertical stress in soil.

A surcharge that extends over a large (effectively infinite) loaded area raises the vertical stress at every depth below it by an amount equal to the surcharge pressure, rather than changing the shape of the profile below the surface. A surcharge from a finite-sized footing instead attenuates with depth and spreads laterally, so it takes a method beyond this chapter’s scope (such as the Boussinesq approach) to find the added stress at a given depth.

Example: vertical effective stress with surcharge and water table

A soil profile has a surcharge of $q_{s} = 50$ kPa applied at the surface. The soil above the water table (depth $0$ - $2$ m) has a unit weight of $γ = 18$ kN/m³, and below the water table (depth $2$ - $5$ m) has a saturated unit weight of $γ_{s a t} = 20$ kN/m³. Find the vertical effective stress at a depth of $5$ m.

Total vertical stress: $σ = q_{s} + (18) (2) + (20) (3) = 50 + 36 + 60 = 146$ kPa

Pore water pressure at 5 m (3 m below the water table): $u = (9.81) (3) = 29.4$ kPa

Effective stress: $σ^{'} = σ - u = 146 - 29.4 = 116.6$ kPa

Answer: $σ^{'} \approx 116.6$ kPa

Horizontal stress profiles and forces

Definitions

Earth pressure: Earth pressure refers to the lateral pressure exerted by soil on retaining structures.

Earth pressure depends on how the soil mass is allowed to deform:

Active conditions occur when the wall moves away from the soil enough for the soil to expand laterally.
Passive conditions occur when the wall moves into the soil enough to compress the soil laterally.
At-rest conditions occur when the wall does not move enough to mobilize active or passive states.

Active earth pressure coefficient (Rankine):

$K_{a} = tan^{2} (4 5^{\circ} - \frac{ϕ}{2})$

Passive earth pressure coefficient (Rankine):

$K_{p} = tan^{2} (4 5^{\circ} + \frac{ϕ}{2})$

At-rest earth pressure coefficient:

For normally consolidated soil:

$K_{0} = 1 - sin ϕ$

For overconsolidated soil:

$K_{0} = (1 - sin ϕ) \cdot OCR^{s i n ϕ}$

Example: active thrust on a wall

A vertical wall retains $H = 4$ m of dry sand with $γ = 17$ kN/m³, $ϕ = 30°$ , and no cohesion. Find the total active thrust per unit length of wall.

Active earth pressure coefficient: $K_{a} = tan^{2} (45° - 15°) = 0.333$

Active pressure at the base: $σ_{a} = K_{a} γ H = (0.333) (17) (4) = 22.6$ kPa

Total active thrust (area of the triangular pressure distribution): $P_{a} = \frac{1}{2} K_{a} γ H^{2} = \frac{1}{2} (0.333) (17) (4)^{2} = 45.3$ kN/m

Answer: $P_{a} \approx 45.3$ kN/m

Exam tip: The earth-pressure coefficients above, along with the retaining-wall factor-of-safety equations and the slope-stability equations later in this chapter, are all given in the Geotechnical section of the FE Reference Handbook. The exam tests your ability to locate and apply the correct equation, not to recall it from memory.

Retaining walls

Definitions

Retaining walls: Retaining walls are structures designed to resist lateral earth pressure and retain soil at different elevations.

Retaining wall design checks typically focus on overturning, sliding, and bearing capacity.

Overturning:

$F S_{overturning} = \frac{\sum M _{R}}{M _{O}}$

Sliding:

$F S_{sliding} = \frac{\sum F _{R}}{\sum F _{D}}$

$F S_{sliding} = \frac{( \sum V ) tan δ + B C _{a} + P _{p}}{P _{a} cos α}$

Bearing capacity:

$F S_{bearing capacity} = \frac{q _{UL T}}{q _{toe}}$

Toe stress:

$q_{toe} = \frac{\sum V}{B} (1 + \frac{6 e}{B})$

Eccentricity:

$e = \frac{B}{2} - (\frac{\sum M _{R} - M _{O}}{\sum V})$

Where:

$e$ = eccentricity
$B$ = width of base
$M_{R}$ = resisting moment
$M_{O}$ = overturning moment
$F_{R}$ = resisting forces
$F_{D}$ = driving forces
$V$ = vertical forces
$δ = k_{1} ϕ$ = friction angle between the base of the wall and the soil
$C_{a} = k_{2} c$ = adhesion between the base of the wall and the soil
$k_{1}$ , $k_{2}$ : given, typically range from 1/2 to 2/3

Example: toe stress and bearing capacity factor of safety

A footing has $B = 3$ m, total vertical force 450 kN, resisting moment 900 kN·m, overturning moment 300 kN·m, and soil $q_{u lt} = 250$ kPa. Find the factor of safety against bearing capacity failure.

$e = \frac{3}{2} - \frac{900 - 300}{450} = 0.167 m$

$q_{toe} = \frac{450}{3} (1 + \frac{6 ( 0.167 )}{3}) = 200.0 kPa$

$F S_{bearing capacity} = \frac{250}{200.0} = 1.25$

Answer: $F S_{bearing capacity} = 1.25$

Slope stability

Definitions

Slope stability: Slope stability evaluates the ability of a soil slope to resist failure along a potential slip surface.

A common approach is to compare the available shear resistance along an assumed slip surface to the shear force required for equilibrium.

Factor of safety:

$FS = \frac{T _{FF}}{T _{MOB}}$

Shearing resistance (along slip surface):

$T_{FF} = c L_{s} + W_{M} cos α_{s} tan ϕ$

Mobilized shear force:

$T_{MOB} = W_{M} sin α_{s}$

Where:

$c$ = cohesion
$ϕ$ = angle of internal friction
$L_{s}$ = length of assumed planar slip surface
$W_{M}$ = weight of soil above slip surface
$α_{s}$ = angle of assumed slip surface w.r.t. horizontal

Example: factor of safety against sliding

An excavated slope has an assumed planar slip surface with length $L_{s} = 12$ m and angle $α_{s} = 25°$ from horizontal. The soil above the slip surface weighs $W_{M} = 450$ kN per unit length of slope, has cohesion $c = 15$ kPa, and friction angle $ϕ = 20°$ . Find the factor of safety against sliding.

Shearing resistance: $T_{FF} = c L_{s} + W_{M} cos α_{s} tan ϕ = (15) (12) + (450) (cos 25°) (tan 20°) = 180 + 148.4 = 328.4$ kN

Mobilized shear force: $T_{MOB} = W_{M} sin α_{s} = (450) (sin 25°) = 190.2$ kN

Factor of safety: $FS = \frac{T _{FF}}{T _{MOB}} = \frac{328.4}{190.2} = 1.73$

Answer: $FS \approx 1.73$

Bearing capacity

Maximum soil pressure foundation can support without failure or excessive settlement
Ultimate bearing capacity formula:
- $q_{u lt} = c N_{c} + γ^{'} D_{f} N_{q} + \frac{1}{2} γ^{'} B N_{γ}$
Key factors: $N_{c}$ , $N_{q}$ , $N_{γ}$ , $D_{f}$ , $B$

Vertical stress profiles

Vertical stress increases with depth due to overlying soil and loads
Defined as stress perpendicular to horizontal soil plane

Vertical stress profiles with surcharge

Surcharge: additional surface load (e.g., traffic, structures)
Surcharge increases vertical stress and shifts stress profile upward

Horizontal stress profiles and forces

Earth pressure: lateral pressure soil exerts on structures
Types of earth pressure:
- Active: wall moves away, soil expands
- Passive: wall moves into soil, soil compresses
- At-rest: no significant wall movement
Key coefficients:
- $K_{a} = tan^{2} (4 5^{\circ} - \frac{ϕ}{2})$ (active)
- $K_{p} = tan^{2} (4 5^{\circ} + \frac{ϕ}{2})$ (passive)
- $K_{0} = 1 - sin ϕ$ (at-rest, normally consolidated)
- $K_{0} = (1 - sin ϕ) \cdot OCR^{s i n ϕ}$ (at-rest, overconsolidated)

Retaining walls

Structures to resist lateral earth pressure and retain soil
Design checks:
- Overturning: $F S_{overturning} = \frac{\sum M _{R}}{M _{O}}$
- Sliding: $F S_{sliding} = \frac{\sum F _{R}}{\sum F _{D}}$ or alternate formula
- Bearing capacity: $F S_{bearing capacity} = \frac{q _{UL T}}{q _{toe}}$
Toe stress and eccentricity formulas:
- $q_{toe} = \frac{\sum V}{B} (1 + \frac{6 e}{B})$
- $e = \frac{B}{2} - (\frac{\sum M _{R} - M _{O}}{\sum V})$

Slope stability

Assesses soil slope’s resistance to failure along slip surface
Factor of Safety: $FS = \frac{T _{FF}}{T _{MOB}}$
- $T_{FF} = c L_{s} + W_{M} cos α_{s} tan ϕ$ (shearing resistance)
- $T_{MOB} = W_{M} sin α_{s}$ (mobilized shear force)
Key terms: cohesion ( $c$ ), friction angle ( $ϕ$ ), slip surface length ( $L_{s}$ ), soil weight ( $W_{M}$ ), slip angle ( $α_{s}$ )

Next | 10.4 Soil classification

Bearing capacity, stress and slope stability

This chapter covers the following topics:

Bearing capacity
Vertical stress profiles
Vertical stress profiles with surcharge
Horizontal stress profiles and forces
Retaining walls
Slope stability

Bearing capacity

Definitions

Bearing capacity: Bearing capacity is the maximum pressure that soil can safely support from a foundation without experiencing shear failure or excessive settlement.

A common expression for the ultimate bearing capacity of a strip footing is:

$q_{u lt} = c N_{c} + γ^{'} D_{f} N_{q} + \frac{1}{2} γ^{'} B N_{γ}$

Where:

$c$ = cohesion of the soil
$γ^{'}$ = effective unit weight of the soil
$N_{c}$ , $N_{q}$ , $N_{γ}$ = bearing capacity factors, which depend on $ϕ$ . The FE Reference Handbook defines them but does not list their values, so a problem supplies them. The expression $N_{q} = e^{π t a n ϕ} tan^{2} (45° + \frac{ϕ}{2})$ (the form used in Meyerhof’s and Vesic’s methods, not Terzaghi’s) shows how $N_{q}$ follows from $ϕ$ .
$D_{f}$ = depth of footing below ground surface
$B$ = width of strip footing

Example: ultimate bearing capacity

A strip footing has $B = 2$ m, $D_{f} = 1.5$ m, $c = 20$ kPa, and $γ^{'} = 18$ kN/m³. For $ϕ = 30°$ , Meyerhof’s bearing capacity factors are $N_{c} = 30.14$ , $N_{q} = 18.4$ , $N_{γ} = 15.67$ .

$q_{u lt} = (20) (30.14) + (18) (1.5) (18.4) + \frac{1}{2} (18) (2) (15.67) = 1, 381.7 kPa$

Answer: $q_{u lt} \approx 1, 381.7$ kPa

Vertical stress profiles

Definitions

Vertical stress: Vertical stress is the stress acting perpendicular to a horizontal plane in soil due to self-weight and applied loads.

Vertical stress generally increases with depth because more soil (and any applied loads) lies above the point of interest.

Vertical stress profiles with surcharge

Definitions

Surcharge: Surcharge is an additional surface load applied to the ground, such as traffic load or structural loading, that increases vertical stress in soil.

Example: vertical effective stress with surcharge and water table

A soil profile has a surcharge of $q_{s} = 50$ kPa applied at the surface. The soil above the water table (depth $0$ - $2$ m) has a unit weight of $γ = 18$ kN/m³, and below the water table (depth $2$ - $5$ m) has a saturated unit weight of $γ_{s a t} = 20$ kN/m³. Find the vertical effective stress at a depth of $5$ m.

Total vertical stress: $σ = q_{s} + (18) (2) + (20) (3) = 50 + 36 + 60 = 146$ kPa

Pore water pressure at 5 m (3 m below the water table): $u = (9.81) (3) = 29.4$ kPa

Effective stress: $σ^{'} = σ - u = 146 - 29.4 = 116.6$ kPa

Answer: $σ^{'} \approx 116.6$ kPa

Horizontal stress profiles and forces

Definitions

Earth pressure: Earth pressure refers to the lateral pressure exerted by soil on retaining structures.

Earth pressure depends on how the soil mass is allowed to deform:

Active conditions occur when the wall moves away from the soil enough for the soil to expand laterally.
Passive conditions occur when the wall moves into the soil enough to compress the soil laterally.
At-rest conditions occur when the wall does not move enough to mobilize active or passive states.

Active earth pressure coefficient (Rankine):

$K_{a} = tan^{2} (4 5^{\circ} - \frac{ϕ}{2})$

Passive earth pressure coefficient (Rankine):

$K_{p} = tan^{2} (4 5^{\circ} + \frac{ϕ}{2})$

At-rest earth pressure coefficient:

For normally consolidated soil:

$K_{0} = 1 - sin ϕ$

For overconsolidated soil:

$K_{0} = (1 - sin ϕ) \cdot OCR^{s i n ϕ}$

Example: active thrust on a wall

A vertical wall retains $H = 4$ m of dry sand with $γ = 17$ kN/m³, $ϕ = 30°$ , and no cohesion. Find the total active thrust per unit length of wall.

Active earth pressure coefficient: $K_{a} = tan^{2} (45° - 15°) = 0.333$

Active pressure at the base: $σ_{a} = K_{a} γ H = (0.333) (17) (4) = 22.6$ kPa

Total active thrust (area of the triangular pressure distribution): $P_{a} = \frac{1}{2} K_{a} γ H^{2} = \frac{1}{2} (0.333) (17) (4)^{2} = 45.3$ kN/m

Answer: $P_{a} \approx 45.3$ kN/m

Retaining walls

Definitions

Retaining walls: Retaining walls are structures designed to resist lateral earth pressure and retain soil at different elevations.

Retaining wall design checks typically focus on overturning, sliding, and bearing capacity.

Overturning:

$F S_{overturning} = \frac{\sum M _{R}}{M _{O}}$

Sliding:

$F S_{sliding} = \frac{\sum F _{R}}{\sum F _{D}}$

$F S_{sliding} = \frac{( \sum V ) tan δ + B C _{a} + P _{p}}{P _{a} cos α}$

Bearing capacity:

$F S_{bearing capacity} = \frac{q _{UL T}}{q _{toe}}$

Toe stress:

$q_{toe} = \frac{\sum V}{B} (1 + \frac{6 e}{B})$

Eccentricity:

$e = \frac{B}{2} - (\frac{\sum M _{R} - M _{O}}{\sum V})$

Where:

$e$ = eccentricity
$B$ = width of base
$M_{R}$ = resisting moment
$M_{O}$ = overturning moment
$F_{R}$ = resisting forces
$F_{D}$ = driving forces
$V$ = vertical forces
$δ = k_{1} ϕ$ = friction angle between the base of the wall and the soil
$C_{a} = k_{2} c$ = adhesion between the base of the wall and the soil
$k_{1}$ , $k_{2}$ : given, typically range from 1/2 to 2/3

Example: toe stress and bearing capacity factor of safety

A footing has $B = 3$ m, total vertical force 450 kN, resisting moment 900 kN·m, overturning moment 300 kN·m, and soil $q_{u lt} = 250$ kPa. Find the factor of safety against bearing capacity failure.

$e = \frac{3}{2} - \frac{900 - 300}{450} = 0.167 m$

$q_{toe} = \frac{450}{3} (1 + \frac{6 ( 0.167 )}{3}) = 200.0 kPa$

$F S_{bearing capacity} = \frac{250}{200.0} = 1.25$

Answer: $F S_{bearing capacity} = 1.25$

Slope stability

Definitions

Slope stability: Slope stability evaluates the ability of a soil slope to resist failure along a potential slip surface.

A common approach is to compare the available shear resistance along an assumed slip surface to the shear force required for equilibrium.

Factor of safety:

$FS = \frac{T _{FF}}{T _{MOB}}$

Shearing resistance (along slip surface):

$T_{FF} = c L_{s} + W_{M} cos α_{s} tan ϕ$

Mobilized shear force:

$T_{MOB} = W_{M} sin α_{s}$

Where:

$c$ = cohesion
$ϕ$ = angle of internal friction
$L_{s}$ = length of assumed planar slip surface
$W_{M}$ = weight of soil above slip surface
$α_{s}$ = angle of assumed slip surface w.r.t. horizontal

Example: factor of safety against sliding

An excavated slope has an assumed planar slip surface with length $L_{s} = 12$ m and angle $α_{s} = 25°$ from horizontal. The soil above the slip surface weighs $W_{M} = 450$ kN per unit length of slope, has cohesion $c = 15$ kPa, and friction angle $ϕ = 20°$ . Find the factor of safety against sliding.

Shearing resistance: $T_{FF} = c L_{s} + W_{M} cos α_{s} tan ϕ = (15) (12) + (450) (cos 25°) (tan 20°) = 180 + 148.4 = 328.4$ kN

Mobilized shear force: $T_{MOB} = W_{M} sin α_{s} = (450) (sin 25°) = 190.2$ kN

Factor of safety: $FS = \frac{T _{FF}}{T _{MOB}} = \frac{328.4}{190.2} = 1.73$

Answer: $FS \approx 1.73$

Achievable FE Civil

10. Soil mechanics

Our FE Civil course is now in "early access" - get 50% off for a limited time.

Bearing capacity, stress and slope stability

8 min read

Font

Discuss

Feedback

This chapter covers the following topics:

Bearing capacity
Vertical stress profiles
Vertical stress profiles with surcharge
Horizontal stress profiles and forces
Retaining walls
Slope stability

Bearing capacity

Definitions

Bearing capacity: Bearing capacity is the maximum pressure that soil can safely support from a foundation without experiencing shear failure or excessive settlement.

A common expression for the ultimate bearing capacity of a strip footing is:

$q_{u lt} = c N_{c} + γ^{'} D_{f} N_{q} + \frac{1}{2} γ^{'} B N_{γ}$

Where:

$c$ = cohesion of the soil
$γ^{'}$ = effective unit weight of the soil
$N_{c}$ , $N_{q}$ , $N_{γ}$ = bearing capacity factors, which depend on $ϕ$ . The FE Reference Handbook defines them but does not list their values, so a problem supplies them. The expression $N_{q} = e^{π t a n ϕ} tan^{2} (45° + \frac{ϕ}{2})$ (the form used in Meyerhof’s and Vesic’s methods, not Terzaghi’s) shows how $N_{q}$ follows from $ϕ$ .
$D_{f}$ = depth of footing below ground surface
$B$ = width of strip footing

Example: ultimate bearing capacity

A strip footing has $B = 2$ m, $D_{f} = 1.5$ m, $c = 20$ kPa, and $γ^{'} = 18$ kN/m³. For $ϕ = 30°$ , Meyerhof’s bearing capacity factors are $N_{c} = 30.14$ , $N_{q} = 18.4$ , $N_{γ} = 15.67$ .

$q_{u lt} = (20) (30.14) + (18) (1.5) (18.4) + \frac{1}{2} (18) (2) (15.67) = 1, 381.7 kPa$

Answer: $q_{u lt} \approx 1, 381.7$ kPa

Vertical stress profiles

Definitions

Vertical stress: Vertical stress is the stress acting perpendicular to a horizontal plane in soil due to self-weight and applied loads.

Vertical stress generally increases with depth because more soil (and any applied loads) lies above the point of interest.

Vertical stress profiles with surcharge

Definitions

Surcharge: Surcharge is an additional surface load applied to the ground, such as traffic load or structural loading, that increases vertical stress in soil.

Example: vertical effective stress with surcharge and water table

A soil profile has a surcharge of $q_{s} = 50$ kPa applied at the surface. The soil above the water table (depth $0$ - $2$ m) has a unit weight of $γ = 18$ kN/m³, and below the water table (depth $2$ - $5$ m) has a saturated unit weight of $γ_{s a t} = 20$ kN/m³. Find the vertical effective stress at a depth of $5$ m.

Total vertical stress: $σ = q_{s} + (18) (2) + (20) (3) = 50 + 36 + 60 = 146$ kPa

Pore water pressure at 5 m (3 m below the water table): $u = (9.81) (3) = 29.4$ kPa

Effective stress: $σ^{'} = σ - u = 146 - 29.4 = 116.6$ kPa

Answer: $σ^{'} \approx 116.6$ kPa

Horizontal stress profiles and forces

Definitions

Earth pressure: Earth pressure refers to the lateral pressure exerted by soil on retaining structures.

Earth pressure depends on how the soil mass is allowed to deform:

Active conditions occur when the wall moves away from the soil enough for the soil to expand laterally.
Passive conditions occur when the wall moves into the soil enough to compress the soil laterally.
At-rest conditions occur when the wall does not move enough to mobilize active or passive states.

Active earth pressure coefficient (Rankine):

$K_{a} = tan^{2} (4 5^{\circ} - \frac{ϕ}{2})$

Passive earth pressure coefficient (Rankine):

$K_{p} = tan^{2} (4 5^{\circ} + \frac{ϕ}{2})$

At-rest earth pressure coefficient:

For normally consolidated soil:

$K_{0} = 1 - sin ϕ$

For overconsolidated soil:

$K_{0} = (1 - sin ϕ) \cdot OCR^{s i n ϕ}$

Example: active thrust on a wall

A vertical wall retains $H = 4$ m of dry sand with $γ = 17$ kN/m³, $ϕ = 30°$ , and no cohesion. Find the total active thrust per unit length of wall.

Active earth pressure coefficient: $K_{a} = tan^{2} (45° - 15°) = 0.333$

Active pressure at the base: $σ_{a} = K_{a} γ H = (0.333) (17) (4) = 22.6$ kPa

Total active thrust (area of the triangular pressure distribution): $P_{a} = \frac{1}{2} K_{a} γ H^{2} = \frac{1}{2} (0.333) (17) (4)^{2} = 45.3$ kN/m

Answer: $P_{a} \approx 45.3$ kN/m

Retaining walls

Definitions

Retaining walls: Retaining walls are structures designed to resist lateral earth pressure and retain soil at different elevations.

Retaining wall design checks typically focus on overturning, sliding, and bearing capacity.

Overturning:

$F S_{overturning} = \frac{\sum M _{R}}{M _{O}}$

Sliding:

$F S_{sliding} = \frac{\sum F _{R}}{\sum F _{D}}$

$F S_{sliding} = \frac{( \sum V ) tan δ + B C _{a} + P _{p}}{P _{a} cos α}$

Bearing capacity:

$F S_{bearing capacity} = \frac{q _{UL T}}{q _{toe}}$

Toe stress:

$q_{toe} = \frac{\sum V}{B} (1 + \frac{6 e}{B})$

Eccentricity:

$e = \frac{B}{2} - (\frac{\sum M _{R} - M _{O}}{\sum V})$

Where:

$e$ = eccentricity
$B$ = width of base
$M_{R}$ = resisting moment
$M_{O}$ = overturning moment
$F_{R}$ = resisting forces
$F_{D}$ = driving forces
$V$ = vertical forces
$δ = k_{1} ϕ$ = friction angle between the base of the wall and the soil
$C_{a} = k_{2} c$ = adhesion between the base of the wall and the soil
$k_{1}$ , $k_{2}$ : given, typically range from 1/2 to 2/3

Example: toe stress and bearing capacity factor of safety

A footing has $B = 3$ m, total vertical force 450 kN, resisting moment 900 kN·m, overturning moment 300 kN·m, and soil $q_{u lt} = 250$ kPa. Find the factor of safety against bearing capacity failure.

$e = \frac{3}{2} - \frac{900 - 300}{450} = 0.167 m$

$q_{toe} = \frac{450}{3} (1 + \frac{6 ( 0.167 )}{3}) = 200.0 kPa$

$F S_{bearing capacity} = \frac{250}{200.0} = 1.25$

Answer: $F S_{bearing capacity} = 1.25$

Slope stability

Definitions

Slope stability: Slope stability evaluates the ability of a soil slope to resist failure along a potential slip surface.

A common approach is to compare the available shear resistance along an assumed slip surface to the shear force required for equilibrium.

Factor of safety:

$FS = \frac{T _{FF}}{T _{MOB}}$

Shearing resistance (along slip surface):

$T_{FF} = c L_{s} + W_{M} cos α_{s} tan ϕ$

Mobilized shear force:

$T_{MOB} = W_{M} sin α_{s}$

Where:

$c$ = cohesion
$ϕ$ = angle of internal friction
$L_{s}$ = length of assumed planar slip surface
$W_{M}$ = weight of soil above slip surface
$α_{s}$ = angle of assumed slip surface w.r.t. horizontal

Example: factor of safety against sliding

An excavated slope has an assumed planar slip surface with length $L_{s} = 12$ m and angle $α_{s} = 25°$ from horizontal. The soil above the slip surface weighs $W_{M} = 450$ kN per unit length of slope, has cohesion $c = 15$ kPa, and friction angle $ϕ = 20°$ . Find the factor of safety against sliding.

Shearing resistance: $T_{FF} = c L_{s} + W_{M} cos α_{s} tan ϕ = (15) (12) + (450) (cos 25°) (tan 20°) = 180 + 148.4 = 328.4$ kN

Mobilized shear force: $T_{MOB} = W_{M} sin α_{s} = (450) (sin 25°) = 190.2$ kN

Factor of safety: $FS = \frac{T _{FF}}{T _{MOB}} = \frac{328.4}{190.2} = 1.73$

Answer: $FS \approx 1.73$

Bearing capacity

Maximum soil pressure foundation can support without failure or excessive settlement
Ultimate bearing capacity formula:
- $q_{u lt} = c N_{c} + γ^{'} D_{f} N_{q} + \frac{1}{2} γ^{'} B N_{γ}$
Key factors: $N_{c}$ , $N_{q}$ , $N_{γ}$ , $D_{f}$ , $B$

Vertical stress profiles

Vertical stress increases with depth due to overlying soil and loads
Defined as stress perpendicular to horizontal soil plane

Vertical stress profiles with surcharge

Surcharge: additional surface load (e.g., traffic, structures)
Surcharge increases vertical stress and shifts stress profile upward

Horizontal stress profiles and forces

Earth pressure: lateral pressure soil exerts on structures
Types of earth pressure:
- Active: wall moves away, soil expands
- Passive: wall moves into soil, soil compresses
- At-rest: no significant wall movement
Key coefficients:
- $K_{a} = tan^{2} (4 5^{\circ} - \frac{ϕ}{2})$ (active)
- $K_{p} = tan^{2} (4 5^{\circ} + \frac{ϕ}{2})$ (passive)
- $K_{0} = 1 - sin ϕ$ (at-rest, normally consolidated)
- $K_{0} = (1 - sin ϕ) \cdot OCR^{s i n ϕ}$ (at-rest, overconsolidated)

Retaining walls

Structures to resist lateral earth pressure and retain soil
Design checks:
- Overturning: $F S_{overturning} = \frac{\sum M _{R}}{M _{O}}$
- Sliding: $F S_{sliding} = \frac{\sum F _{R}}{\sum F _{D}}$ or alternate formula
- Bearing capacity: $F S_{bearing capacity} = \frac{q _{UL T}}{q _{toe}}$
Toe stress and eccentricity formulas:
- $q_{toe} = \frac{\sum V}{B} (1 + \frac{6 e}{B})$
- $e = \frac{B}{2} - (\frac{\sum M _{R} - M _{O}}{\sum V})$

Slope stability

Assesses soil slope’s resistance to failure along slip surface
Factor of Safety: $FS = \frac{T _{FF}}{T _{MOB}}$
- $T_{FF} = c L_{s} + W_{M} cos α_{s} tan ϕ$ (shearing resistance)
- $T_{MOB} = W_{M} sin α_{s}$ (mobilized shear force)
Key terms: cohesion ( $c$ ), friction angle ( $ϕ$ ), slip surface length ( $L_{s}$ ), soil weight ( $W_{M}$ ), slip angle ( $α_{s}$ )

Next | 10.4 Soil classification

Bearing capacity, stress and slope stability

This chapter covers the following topics:

Bearing capacity
Vertical stress profiles
Vertical stress profiles with surcharge
Horizontal stress profiles and forces
Retaining walls
Slope stability

Bearing capacity

Definitions

Bearing capacity: Bearing capacity is the maximum pressure that soil can safely support from a foundation without experiencing shear failure or excessive settlement.

A common expression for the ultimate bearing capacity of a strip footing is:

$q_{u lt} = c N_{c} + γ^{'} D_{f} N_{q} + \frac{1}{2} γ^{'} B N_{γ}$

Where:

$c$ = cohesion of the soil
$γ^{'}$ = effective unit weight of the soil
$N_{c}$ , $N_{q}$ , $N_{γ}$ = bearing capacity factors, which depend on $ϕ$ . The FE Reference Handbook defines them but does not list their values, so a problem supplies them. The expression $N_{q} = e^{π t a n ϕ} tan^{2} (45° + \frac{ϕ}{2})$ (the form used in Meyerhof’s and Vesic’s methods, not Terzaghi’s) shows how $N_{q}$ follows from $ϕ$ .
$D_{f}$ = depth of footing below ground surface
$B$ = width of strip footing

Example: ultimate bearing capacity

A strip footing has $B = 2$ m, $D_{f} = 1.5$ m, $c = 20$ kPa, and $γ^{'} = 18$ kN/m³. For $ϕ = 30°$ , Meyerhof’s bearing capacity factors are $N_{c} = 30.14$ , $N_{q} = 18.4$ , $N_{γ} = 15.67$ .

$q_{u lt} = (20) (30.14) + (18) (1.5) (18.4) + \frac{1}{2} (18) (2) (15.67) = 1, 381.7 kPa$

Answer: $q_{u lt} \approx 1, 381.7$ kPa

Vertical stress profiles

Definitions

Vertical stress: Vertical stress is the stress acting perpendicular to a horizontal plane in soil due to self-weight and applied loads.

Vertical stress generally increases with depth because more soil (and any applied loads) lies above the point of interest.

Vertical stress profiles with surcharge

Definitions

Surcharge: Surcharge is an additional surface load applied to the ground, such as traffic load or structural loading, that increases vertical stress in soil.

Example: vertical effective stress with surcharge and water table

A soil profile has a surcharge of $q_{s} = 50$ kPa applied at the surface. The soil above the water table (depth $0$ - $2$ m) has a unit weight of $γ = 18$ kN/m³, and below the water table (depth $2$ - $5$ m) has a saturated unit weight of $γ_{s a t} = 20$ kN/m³. Find the vertical effective stress at a depth of $5$ m.

Total vertical stress: $σ = q_{s} + (18) (2) + (20) (3) = 50 + 36 + 60 = 146$ kPa

Pore water pressure at 5 m (3 m below the water table): $u = (9.81) (3) = 29.4$ kPa

Effective stress: $σ^{'} = σ - u = 146 - 29.4 = 116.6$ kPa

Answer: $σ^{'} \approx 116.6$ kPa

Horizontal stress profiles and forces

Definitions

Earth pressure: Earth pressure refers to the lateral pressure exerted by soil on retaining structures.

Earth pressure depends on how the soil mass is allowed to deform:

Active conditions occur when the wall moves away from the soil enough for the soil to expand laterally.
Passive conditions occur when the wall moves into the soil enough to compress the soil laterally.
At-rest conditions occur when the wall does not move enough to mobilize active or passive states.

Active earth pressure coefficient (Rankine):

$K_{a} = tan^{2} (4 5^{\circ} - \frac{ϕ}{2})$

Passive earth pressure coefficient (Rankine):

$K_{p} = tan^{2} (4 5^{\circ} + \frac{ϕ}{2})$

At-rest earth pressure coefficient:

For normally consolidated soil:

$K_{0} = 1 - sin ϕ$

For overconsolidated soil:

$K_{0} = (1 - sin ϕ) \cdot OCR^{s i n ϕ}$

Example: active thrust on a wall

A vertical wall retains $H = 4$ m of dry sand with $γ = 17$ kN/m³, $ϕ = 30°$ , and no cohesion. Find the total active thrust per unit length of wall.

Active earth pressure coefficient: $K_{a} = tan^{2} (45° - 15°) = 0.333$

Active pressure at the base: $σ_{a} = K_{a} γ H = (0.333) (17) (4) = 22.6$ kPa

Total active thrust (area of the triangular pressure distribution): $P_{a} = \frac{1}{2} K_{a} γ H^{2} = \frac{1}{2} (0.333) (17) (4)^{2} = 45.3$ kN/m

Answer: $P_{a} \approx 45.3$ kN/m

Retaining walls

Definitions

Retaining walls: Retaining walls are structures designed to resist lateral earth pressure and retain soil at different elevations.

Retaining wall design checks typically focus on overturning, sliding, and bearing capacity.

Overturning:

$F S_{overturning} = \frac{\sum M _{R}}{M _{O}}$

Sliding:

$F S_{sliding} = \frac{\sum F _{R}}{\sum F _{D}}$

$F S_{sliding} = \frac{( \sum V ) tan δ + B C _{a} + P _{p}}{P _{a} cos α}$

Bearing capacity:

$F S_{bearing capacity} = \frac{q _{UL T}}{q _{toe}}$

Toe stress:

$q_{toe} = \frac{\sum V}{B} (1 + \frac{6 e}{B})$

Eccentricity:

$e = \frac{B}{2} - (\frac{\sum M _{R} - M _{O}}{\sum V})$

Where:

$e$ = eccentricity
$B$ = width of base
$M_{R}$ = resisting moment
$M_{O}$ = overturning moment
$F_{R}$ = resisting forces
$F_{D}$ = driving forces
$V$ = vertical forces
$δ = k_{1} ϕ$ = friction angle between the base of the wall and the soil
$C_{a} = k_{2} c$ = adhesion between the base of the wall and the soil
$k_{1}$ , $k_{2}$ : given, typically range from 1/2 to 2/3

Example: toe stress and bearing capacity factor of safety

A footing has $B = 3$ m, total vertical force 450 kN, resisting moment 900 kN·m, overturning moment 300 kN·m, and soil $q_{u lt} = 250$ kPa. Find the factor of safety against bearing capacity failure.

$e = \frac{3}{2} - \frac{900 - 300}{450} = 0.167 m$

$q_{toe} = \frac{450}{3} (1 + \frac{6 ( 0.167 )}{3}) = 200.0 kPa$

$F S_{bearing capacity} = \frac{250}{200.0} = 1.25$

Answer: $F S_{bearing capacity} = 1.25$

Slope stability

Definitions

Slope stability: Slope stability evaluates the ability of a soil slope to resist failure along a potential slip surface.

A common approach is to compare the available shear resistance along an assumed slip surface to the shear force required for equilibrium.

Factor of safety:

$FS = \frac{T _{FF}}{T _{MOB}}$

Shearing resistance (along slip surface):

$T_{FF} = c L_{s} + W_{M} cos α_{s} tan ϕ$

Mobilized shear force:

$T_{MOB} = W_{M} sin α_{s}$

Where:

$c$ = cohesion
$ϕ$ = angle of internal friction
$L_{s}$ = length of assumed planar slip surface
$W_{M}$ = weight of soil above slip surface
$α_{s}$ = angle of assumed slip surface w.r.t. horizontal

Example: factor of safety against sliding

An excavated slope has an assumed planar slip surface with length $L_{s} = 12$ m and angle $α_{s} = 25°$ from horizontal. The soil above the slip surface weighs $W_{M} = 450$ kN per unit length of slope, has cohesion $c = 15$ kPa, and friction angle $ϕ = 20°$ . Find the factor of safety against sliding.

Shearing resistance: $T_{FF} = c L_{s} + W_{M} cos α_{s} tan ϕ = (15) (12) + (450) (cos 25°) (tan 20°) = 180 + 148.4 = 328.4$ kN

Mobilized shear force: $T_{MOB} = W_{M} sin α_{s} = (450) (sin 25°) = 190.2$ kN

Factor of safety: $FS = \frac{T _{FF}}{T _{MOB}} = \frac{328.4}{190.2} = 1.73$

Answer: $FS \approx 1.73$