SME
Soil Mechanics
Bearing capacity
Use this for “A 2 m strip footing sits 1.5 m down in sand with φ′ = 30°, c′ = 5 kPa and γ = 18 kN/m³. Find qᵤ and qₐₗₗ with a factor of safety of 3.”
Diagram
Nq = 18.4, Nc = 30.1, Nγ = 22.4 from φ′ = 30°.
A strip footing: qᵤ = c′Nc + γDf·Nq + ½γB·Nγ.
Drawn
to
scale:
Df
B
=
0.75.
The wedges push the soil out and up beside the footing.
This is an example. Type your own values to start a new problem.
qᵤ (kPa)
=
c′ (kPa)
×
Nc
+
γ (kN/m³)
×
Df (m)
×
Nq
+
0.5
×
γ (kN/m³)
×
B (m)
×
Nγ
qₐₗₗ (kPa)
=
qᵤ (kPa)
÷
3
φ′
Friction angle
example · °
Formulas
Nq
=
e
π × tan(φ′)
×
1 + sin(φ′)
1 − sin(φ′)
18.4011
=
e
π × tan(30°)
×
1 + sin(30°)
1 − sin(30°)
Nc
=
Nq − 1
tan(φ′)
30.1396
=
18.4011 − 1
tan(30°)
Nγ = 2 × (Nq + 1) × tan(φ′)
22.4025 = 2 × (18.4011 + 1) × tan(30°)
qᵤ
=
c′
×
Nc
+
γ
×
Df
×
Nq
+
0.5
×
γ
×
B
×
Nγ
1,050.7732 = 5 × 30.1396 + 18 × 1.5 × 18.4011 + 0.5 × 18 × 2 × 22.4025
qₐₗₗ
=
qᵤ
3
350.2577
=
1,050.7732
3
Assumptions
•
General shear failure: a wedge under the footing pushes the soil beside it out and up.
•
Vesić’s factors: Nq and Nc are Reissner–Prandtl’s; Terzaghi’s table gives larger Nc and Nq, so use your course’s.
•
A clay loaded undrained has φ′ = 0: Nc = π + 2 = 5.14, Nq = 1, Nγ = 0, and c′ is sᵤ.
•
A long strip footing with no water table near its base; qₐₗₗ = qᵤ ÷ 3 (factor of safety 3).
Step-by-step
WE KNOW
φ′ = 30°, c′ = 5 kPa, γ = 18 kN/m³, B = 2 m, Df = 1.5 m
FIND
factor Nq (Nq), factor Nc (Nc), factor Nγ (Nγ), ultimate capacity (qᵤ), allowable pressure (qₐₗₗ)
Step 1 · Find factor Nq (Nq)
Use Nq = eπ × tan(φ′) × (1 + sin(φ′)) ÷ (1 − sin(φ′))
The
soil
beside
the
footing,
carried
round
the
log-spiral
zone;
tan²(45°
+
φ′
2
)
=
1 + sin φ′
1 − sin φ′
.
Nq
=
e
π × tan(30°)
×
1 + sin(30°)
1 − sin(30°)
Nq
=
e
0.57735 × π
×
1 + 0.5
1 − 0.5
Nq
=
e
1.8138
×
1.5
0.5
Nq
=
6.1337
×
1.5
0.5
Nq
=
9.2005
0.5
Nq = 18.4011
Step 2 · Find factor Nc (Nc)
Use Nc = (Nq − 1) ÷ tan(φ′)
The cohesion factor follows from Nq.
φ′ = 30° > 0: cot φ′ is defined
Nc
=
18.4011 − 1
tan(30°)
Nc
=
17.401
0.57735
Nc = 30.1396
Step 3 · Find factor Nγ (Nγ)
Use Nγ = 2 × (Nq + 1) × tan(φ′)
Vesić’s factor for the weight of the soil in the wedges (0 when φ′ = 0).
Nγ = 2 × (18.4011 + 1) × tan(30°)
Nγ = 2 × 19.401 × 0.57735
Nγ = 22.4025
Step 4 · Find ultimate capacity (qᵤ)
Use qᵤ = c′ × Nc + γ × Df × Nq + 0.5 × γ × B × Nγ
Add the three terms: cohesion, the soil above the base, and the soil’s weight under it.
qᵤ = 5 × 30.1396 + 18 × 1.5 × 18.4011 + 0.5 × 18 × 2 × 22.4025
qᵤ = 150.698 + 27 × 18.4011 + 9 × 2 × 22.4025
qᵤ = 150.698 + 496.8297 + 18 × 22.4025
qᵤ = 150.698 + 496.8297 + 403.245
qᵤ = 1,050.773 kPa
Step 5 · Find allowable pressure (qₐₗₗ)
Use qₐₗₗ = qᵤ ÷ 3
Divide by the factor of safety, 3.
qₐₗₗ
=
1,050.773
3
qₐₗₗ = 350.2577 kPa
Check
Put the numbers back in. Both sides should match.
18.4011
=
e
π × tan(30°)
×
1 + sin(30°)
1 − sin(30°)
✓
30.1396
=
18.4011 − 1
tan(30°)
✓
22.4025 = 2 × (18.4011 + 1) × tan(30°) ✓
1,050.773 = 5 × 30.1396 + 18 × 1.5 × 18.4011 + 0.5 × 18 × 2 × 22.4025 ✓
350.2577
=
1,050.773
3
✓
Topic check
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