5. Bearing Capacity of
Shallow Footings
CIV4249: Foundation Engineering
Monash University
Bearing Capacity
• Ultimate or serviceability limit state?
• “What is the maximum pressure which the
soils can withstand for a given foundation
before the soil will fail?”
• Design for less but how much less?
• Uncertainty with respect to:
– Loads
– Capacity
Limit State Design
• Limit state design equation:
yF<fR
• F = action (kN or kPa)
• y = load factor
–
–
–
–
(AS1170) - Loading Code
Dead Load = 1.25
Live Load = 1.50
Hydrostatic = 1.00
• Typical value
– 2/3 dead + 1/3 live
– y = 1.33
• R = capacity (kN or kPa)
• f = capacity redn factor
–
–
–
–
(AS2159) - Piling Code
Static test = 0.70 to 0.90
CPT design = 0.45 to 0.65
SPT design = 0.40 to 0.55
• Why a range?
– variability in site conditions
and in quality or quantity of
exploration
Factor of Safety
• Working or Allowable stress method is currently
used in practice
• No Australian Standard
• By convention Factor of Safety = 2.5 to 3.0
• qallow = qult  FoS
• I want you to apply limit state design principles
• Equivalent “Factor of safety” =
y/f
• For y = 1.33 implies f =
0.44 to 0.53
Geotechnical Design
yF<fR
• Generally working with stresses
• On LHS concerned only with that applied
stress which acts to cause rupture
• On RHS concerned with the available
strength which acts to prevent rupture
Applied Stress, F
• Fwall = 120 kN/m : Wwall = 20 kN/m : Wfoot = 10 kN/m
•
What is the applied stress in these two situations?
1.2m
1.0m
Net Applied Stress, F
• Fwall = 120 kN/m : Wwall = 20 kN/m : Wfoot = 10 kN/m : g = 20 kN/m3
•
What is the net applied stress in these two situations?
1.2m
qnet = 125 kPa
1.0m
qnet = 105 kPa
Net Applied Stress Rule
• For bearing capacity:
qnet applied = s 'below - s 'beside
ALWAYS WORK WITH NET APPLIED STRESSES
NEVER WORK WITH GROSS APPLIED STRESSES
Available Strength, R
• Methods that can be used to determine
available strength:
1. Historical / experience :
• Building Codes may specify allowable values in
particular formations
2. Field loading tests
• Plate loading tests for very large projects
3. Analytical solutions
• Upper and lower bound solutions for special cases
4. Approximate solutions
• Solutions for general cases
Field (Plate) Loading Tests
-
+
0.3m
1.2m
• Testing footing under
actual soil conditions
• Measure load-deflection
behaviour
• Expensive mobilization
and testing
• Need to apply scaling laws
• Different zone of influence
• Affected by fabric – fissuring, partings etc.
Analytical Solutions
• The failure of real soils with weight,
cohesion and friction is a complex
phenomenon, not amenable to simple
theoretical solutions.
• If simplifying assumptions are made, it is
possible to develop particular analytical
solutions.
• These analytical solutions must be based
either on principles of equilibrium or
kinematic admissibility.
Lower Bound Solution
u = 4c
• “If an equilibrium distribution of qstresses
can be found which balances the applied
load, and nowhere violates the yield
criterion, the soil mass will not fail or will
1
2 - i.e. it will
be just at the point of failure”
Weightless
be a lower-bound
estimate of capacity.
soil f = 0
0
2c
2c
4c
Upper Bound Solution
• “If a solution is kinematically admissible
qu r. r/2 = p r.c.r
and simultaneously satisfies equilibrium
qu = 2pc
considerations, failure must result - i.e. it
O estimate of
will be an upper-bound
capacity.”
c
r
c
Weightless
soil f = 0
e.g. slope stability - optimize failure surface; choose FoS
Other classic analytical
solutions for weightless soils:
• Solutions with f = 0 :
– Prandtl smooth punch : qu = 5.14c
– Prandtl rough punch : qu = 5.7c
• Solutions with f  0 :
– Rough punch
passive
active
log spiral
Solutions for real soils
• There is no rigorous mathematical
solution for a soil which contains
cohesion, c, and angle of friction, f, and
weight, g.
• Empirical or numerical approaches must
be used to provide methods of estimating
bearing capacity in practical situations.
• Numerical approaches include finite
element and boundary element methods
and would rarely be used in practice*
Terzaghi Approximate Analysis
• Solution for soil with c, f, g and D > 0
• Solution is based on superposition of 3
separate analytical cases:
– Soil with f and g but c = D = 0 : qu = Ng.f(g)
– Soil with f and D but c = g = 0 : qu = Nq f(D)
– Soil with f and c but g = D = 0 : qu = Nc f(c)
• Each case has a different failure surface,
so superposition is not theoretically valid.
Terzaghi Bearing Equation
qu nett = c.Nc + p'o (Nq - 1) + 0.5Bg'Ng
Solution for c and f only soil
Solution for D and f only soil
Solution for g and f only soil
Terzaghi Bearing Equation
qu nett = c.Nc + p'o (Nq - 1) + 0.5Bg'Ng
Overburden
p'o = g'o D
B
Failure Zone (depth  2B)
Generalized soil strength : c, f Soil unit weight : g' (total or
effective as applicable)
(drainage as applicable)
Adopt weighted average values !
Terzaghi Bearing Equation
qu nett = c.Nc + p'o (Nq - 1) + 0.5Bg'Ng
– applies to strip footing
– Nc, Nq and Ng are functions of f, and are
usually given in graphical form
– c, f and g' refer to soil properties in the failure
zone below the footing
– p'o is the effective overburden pressure at the
founding level
– shear strength contribution above footing level
is ignored : conservative for deeper footings
Application to other than strip
footings
• Strip footings represent a plane-strain
case
• What is different for a rectangular footing?
• Correction factors applied - e.g. Schultz:
– Nc multiplier is (1+ 0.2B/L)
Example #1
1.0
1.7 x 2.3
Stiff Clay : cu = 75 kPa fu = 0o g = 18 kN/m3
•
•
•
•
•
•
•
Shape Factor = (1 + 0.2*1.7/2.3) = 1.148
c = 75 kPa
Nc = 5.7
Nq = 1.0
Ng = undefined
Qu nett = 1.148*75*5.7*1.7*2.3 = 1919 kN
f / y Qu nett = 0.45 * 1919/1.33 = 649 kN = 166 kPa
Example #2
1.0
1.7 x 2.3
Medium Sand : c = 0 kPa f' = 35o g = 20 kN/m3
• c = 0 kPa
• p'o = 1.0*20 = 20 kPa
• Nq = 40
• g' = 10.2kN/m3
• Ng = 40
• Qu nett = [20*(40-1)+0.5*0.852*1.7*10.2*40]*1.7*2.3 = 4205 kN
• f/y Qu nett = 0.45 * 4205/1.33 = 1422 kN = 364 kPa
Footings with eccentric loads
P
e
e < B/6 :
rigid
qmin
qmin = P (1-6e/B)/BL
qmax = P (1+6e/B)/BL
qmax
Footings with eccentric loads
P
e
e > B/6 :
rigid
qmin
qmin = 0
qmax =
4P .
3L(B-2e)
qmax
Meyerhof Method for eccentric
loads
L
e
B
P
2e
L' = L- 2e
Meyerhof Method for eccentric
loads
2.50
PB/L
2.00
q(min)
Meyerhof
q(max)
average
1.50
1.00
0.50
0.00
0.00
0.05
0.10
0.15
0.20
e/B
2-way eccentricity
L
2e2
B
e2
P
2e1
L' = L- 2e1
B' = B- 2e2
e1
Footings with moments
M
P
e
P
e=M
P
treat as equivalent eccentric load
Equivalent footing example
Light tower
5.3x5.3 m
Vertical Load = 500 kN
Equiv Horizontal Load = 30 kN @ 13m above base
Determine:
a)
b)
Maximum and minimum stresses under the footing
Equivalent footing dimensions
Equivalent footing example
Light tower
5.3x5.3 m
•
•
•
•
•
Effective eccentricity = 30*13/500 = 0.78m
e/B = 0.78/5.3 = 0.147 < 0.166B
smin = 500*(1- 6*0.147)/5.32 = 2.1 kPa
smax = 500*(1+ 6*0.147)/5.32 = 33.5 kPa
Effective area = 5.3 * (5.3 - 2*0.78) = 5.3 * 3.74m
Inclined Loads
Fc = Fq = (1 - d / 90)2
Fg = (1 - d / f)2
• Correction Factors, Fc , Fq and Fg
empirically determined from experiments
Meyerhof Approx Analysis
• differs from Terzaghi analysis particularly
for buried footings
– soil above footing base provides not only
surcharge but also strength
– more realistic i.e. less conservative
qu = cNcscdcic + qNqsqdqiq + 0.5g'BNgsgdgig
• s, d, and i are shape, depth and load
inclination factors
Analyses by Hansen, Vesic
qu = cNcscdcicgcbc + qNqsqdqiqgqbq +
0.5g'BNgsgdgigggbg
Nc ,Nq ,Ng : Meyerhof bearing capacity factors
sc ,sq ,sg : shape factors
dc ,dq ,dg : depth factors
ic ,iq ,ig : load inclination factors
gc ,gq ,gg : ground inclination factors
bc ,bq ,bg : base inclination factors
Example 4 - Bearing capacity after
Hansen
Ground inclination = 3.5o
Load inclination = 10o
1.0
Firm Clay : cu = 40 kPa fu = 0o g = 17 kN/m3
1.7 x 2.3
1.5
Medium sand : f' = 34o g = 20 kN/m3
Grading dense : f' = 40o g = 21.5 kN/m3
Determine the ultimate bearing capacity (in kN)
0.6
Example 4 - Bearing capacity after
Hansen
• f = 1.5*34+1.9*40/3.4 =
37o
• Nc = 0
• Nq = 42.9
• Ng = 47.4
• sq = 1.445
• sg = 0.704
• dq = 1.239
• dg = 1.00
•
•
•
•
•
•
•
•
iq = 0.831 (1 = 2)
ig = 0.674 (2 = 3)
gq = 0.80
gg = 0.80
bq = 1.00
bg = 1.00
q = 17*0.6+7*0.4 = 13
g=
1.5*10.2+1.9*11.7/3.4 =
11.0 kN/m3
Qu = 1.7*2.3*(664+168) = 3250 kN
Stratified Deposits - 1
B
soft clay
2B
stiff clay or
dense sand
• For uniform soils, zone of
influence typically ~ 2B
• Failure surface will tend to be
more shallow
• Ignore strength increase?
• Place footing deeper?
trendingof underlying
• Take strength
stiffer stronger
material into account
• Approaches based on taking
weighted average strength
• See Bowles, Das or other text
Stratified Deposits - 2
dense sand
soft clay
• Ignoring underlying layer
unconservative
• compute load spread and
analyze as larger footing
with reduced stress on
underlying soil
• use parameters of
underlying soil in bearing
equation
• again, look at texts for
different approaches
Terminology
• Ultimate Bearing Pressure, qu
– as computed by any number of methods
• Maximum Safe Bearing Pressure, qs
– qs = qu  FoS
• Allowable Bearing Pressure, qa
– take settlement into consideration : qa  qs
• Design Pressure, qd
– construction practicalities/ standardization
may dictate larger footings : qd  qa