SME
Soil Mechanics · Problem type
Square footing
Use this for “A 2 m square footing sits 1.5 m down in sand with φ′ = 30°, c′ = 5 kPa and γ = 18 kN/m³. Find qᵤ and qₐₗₗ.”
Diagram
Nq = 18.4, Nc = 30.1, Nγ = 22.4 from φ′ = 30°.
A
square
footing
(Vesić):
qᵤ
=
c′NcFcs
+
γDf·NqFqs
+
1
2
γB·NγFγs,
Fcs
=
1
+
Nq
Nc
,
Fqs
=
1
+
tan
φ′,
Fγs
=
0.6.
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
×
Fcs
+
γ (kN/m³)
×
Df (m)
×
Nq
×
Fqs
+
0.5
×
γ (kN/m³)
×
B (m)
×
Nγ
×
0.6
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°)
Fcs
=
1
+
Nq
Nc
1.6105
=
1
+
18.4011
30.1396
Fqs = 1 + tan(φ′)
1.5774 = 1 + tan(30°)
qᵤ
=
c′
×
Nc
×
Fcs
+
γ
×
Df
×
Nq
×
Fqs
+
0.5
×
γ
×
B
×
Nγ
×
0.6
1,268.326 = 5 × 30.1396 × 1.6105 + 18 × 1.5 × 18.4011 × 1.5774 + 0.5 × 18 × 2 × 22.4025 × 0.6
qₐₗₗ
=
qᵤ
3
422.7753
=
1,268.326
3
Assumptions
•
General shear failure: a wedge under the footing pushes the soil beside it out and up.
•
Vesić’s method: the strip’s Nq, Nc and Nγ, each term times its shape factor for B = L.
•
Fcs = 1 + Nq ÷ Nc, Fqs = 1 + tan φ′, Fγs = 0.6; depth factors are taken as 1.
•
Terzaghi’s 1.3 and 0.4 go with his own factor table only; no water table near the base; qₐₗₗ = qᵤ ÷ 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), shape factor on c′Nc (Fcs), shape factor on γDfNq (Fqs), 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 = 18.40112
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
=
18.40112 − 1
tan(30°)
Nc
=
17.401
0.57735
Nc = 30.13963
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 × (18.40112 + 1) × tan(30°)
Nγ = 2 × 19.401 × 0.57735
Nγ = 22.40249
Step 4 · Find shape factor on c′Nc (Fcs)
Use Fcs = 1 + Nq ÷ Nc
A
square
base
(
B
L
=
1)
spreads
the
cohesion
term
over
a
wider
zone.
Fcs
=
1
+
18.40112
30.13963
Fcs = 1 + 0.6105
Fcs = 1.61053
Step 5 · Find shape factor on γDfNq (Fqs)
Use Fqs = 1 + tan(φ′)
The soil above the base pushes down on more of the failure zone round a square.
Fqs = 1 + tan(30°)
Fqs = 1 + 0.5774
Fqs = 1.57735
Step 6 · Find ultimate capacity (qᵤ)
Use qᵤ = c′ × Nc × Fcs + γ × Df × Nq × Fqs + 0.5 × γ × B × Nγ × 0.6
Add the three terms, each times its shape factor (Fγs = 1 − 0.4 = 0.6 for a square).
qᵤ = 5 × 30.13963 × 1.61053 + 18 × 1.5 × 18.40112 × 1.57735 + 0.5 × 18 × 2 × 22.40249 × 0.6
qᵤ = 242.7 + 783.68 + 241.95
qᵤ = 1,268.326 kPa
Step 7 · Find allowable pressure (qₐₗₗ)
Use qₐₗₗ = qᵤ ÷ 3
Divide by the factor of safety, 3.
qₐₗₗ
=
1,268.326
3
qₐₗₗ = 422.7753 kPa
Check
Put the numbers back in. Both sides should match.
18.40112
=
e
π × tan(30°)
×
1 + sin(30°)
1 − sin(30°)
✓
30.13963
=
18.40112 − 1
tan(30°)
✓
22.40249 = 2 × (18.40112 + 1) × tan(30°) ✓
1.61053
=
1
+
18.40112
30.13963
✓
1.57735 = 1 + tan(30°) ✓
1,268.326 = 5 × 30.13963 × 1.61053 + 18 × 1.5 × 18.40112 × 1.57735 + 0.5 × 18 × 2 × 22.40249 × 0.6 ✓
422.7753
=
1,268.326
3
✓
Topic check
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