1. 5. Bearing Capacity of Shallow Footings
CIV4249: Foundation Engineering
Monash University

2. 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

3. Limit State Design Limit state design equation: y F < f R
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

4. 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

5. Geotechnical Design y F < f R 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

6. Applied Stress, F Fwall = 120 kN/m : Wwall = 20 kN/m : Wfoot = 10 kN/m
What is the applied stress in these two situations?
1.0m
1.2m
Applied stress = ( ) / 1.2 = 125 kPa in both cases

7. Net Applied Stress, F qnet = 125 kPa qnet = 105 kPa
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?
1.0m
1.2m
Net applied stress for surface footing = 125 kPa
Net applied stress for buried footing = *1.0 = 105 kPa
qnet = 125 kPa
qnet = 105 kPa

8. 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

9. 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

10. 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.

11. 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.

12. Lower Bound Solution
“If an equilibrium distribution of stresses can be found which balances the applied load, and nowhere violates the yield criterion, the soil mass will not fail or will be just at the point of failure” - i.e. it will be a lower-bound estimate of capacity. 1 2
Weightless
soil f = 0 qu
= 4c 2c 2c 4c

13. Upper Bound Solution
Weightless
soil f = 0 qu
“If a solution is kinematically admissible and simultaneously satisfies equilibrium considerations, failure must result - i.e. it will be an upper-bound estimate of capacity.”
qu r. r/2 = p r.c.r
= 2pc r O c
e.g. slope stability - optimize failure surface; choose FoS

14. 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 ¹ 0 :
Rough punch
passive
active
log spiral

15. 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*

16. 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.

17. Terzaghi Bearing Equation
qu nett = c.Nc + p'o (Nq - 1) + 0.5B'N
Solution for c and  only soil
Solution for D and  only soil
Solution for  and  only soil

18. Terzaghi Bearing Equation
qu nett = c.Nc + p'o (Nq - 1) + 0.5B'N
Overburden
p'o = 'o D B
Failure Zone (depth  2B)
Generalized soil strength : c,  (drainage as applicable)
Soil unit weight : ' (total or
effective as applicable)
Adopt weighted average values !

19. Terzaghi Bearing Equation
qu nett = c.Nc + p'o (Nq - 1) + 0.5B'N
applies to strip footing
Nc, Nq and N 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

20. 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)

21. Example #1 1.0 1.7 x 2.3 Stiff Clay : cu = 75 kPa fu = 0o g = 18 kN/m3
( *1.7/2.3) = 1.148
75 kPa
5.7
1.0
undefined
1.148*75*5.7*1.7*2.3 = 1919 kN
0.45 * 1919/1.33 = 649 kN = 166 kPa
Shape Factor =
c =
Nc =
Nq =
Ng =
Qu nett =
 /  Qu nett =

22. Example #2 1.0 1.7 x 2.3 Medium Sand : c = 0 kPa f' = 35o g = 20 kN/m340
10.2kN/m3
[20*(40-1)+0.5*0.852*1.7*10.2*40]*1.7*2.3 = 4205 kN
0.45 * 4205/1.33 = 1422 kN = 364 kPa
c =
p'o =
Nq =
g' =
Ng =
Qu nett =
f/y Qu nett =

23. Footings with eccentric loadsP e
rigid
e < B/6 :
qmin
qmax
qmin = P (1-6e/B)/BL
qmax = P (1+6e/B)/BL

24. Footings with eccentric loadsP e
rigid
e > B/6 :
qmin
qmin = 0
qmax = P L(B-2e)
qmax

25. Meyerhof Method for eccentric loadsB P
L' = L- 2e

26. Meyerhof Method for eccentric loads

27. 2-way eccentricity L B
2e2
B' = B- 2e2 e2 e1 P
2e1
L' = L- 2e1

28. Footings with moments e = M P treat as equivalent eccentric load P M e

29. Equivalent footing example
Light tower
5.3x5.3 m
Vertical Load = 500 kN
Equiv Horizontal Load = 30 13m above base
Determine:
a) Maximum and minimum stresses under the footing
b) Equivalent footing dimensions

30. Equivalent footing example
Light tower
5.3x5.3 m
Effective eccentricity =
e/B =
min =
max =
Effective area =
30*13/500 = 0.78m
0.78/5.3 = < 0.166B
500*(1- 6*0.147)/5.32 = 2.1 kPa
500*(1+ 6*0.147)/5.32 = 33.5 kPa
5.3 * ( *0.78) = 5.3 * 3.74m

31. Inclined Loads Fc = Fq = (1 - d / 90)2 Fg = (1 - d / f)2
Correction Factors, Fc , Fq and Fg empirically determined from experiments

32. 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

33. 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

34. Example 4 - Bearing capacity after Hansen
Ground inclination = 3.5o
Load inclination = 10o
0.6
Firm Clay : cu = 40 kPa fu = 0o g = 17 kN/m3
1.0
1.7 x 2.3
Medium sand : f' = 34o g = 20 kN/m3
1.5
Grading dense : f' = 40o g = 21.5 kN/m3
Determine the ultimate bearing capacity (in kN)

35. Example 4 - Bearing capacity after Hansen
1.5*34+1.9*40/3.4 = 37o
42.9
47.4
1.445
0.704
1.239
1.00
 =
Nc =
Nq =
N =
sq =
s =
dq =
d =
iq =
i =
gq =
g =
bq =
b =
q =
 =
0.831 (1 = 2)
0.674 (2 = 3)
0.80
1.00
17*0.6+7*0.4 = 13
1.5* *11.7/3.4 = 11.0 kN/m3
Qu = 1.7*2.3*( ) = 3250 kN

36. 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?
Take strength of underlying stiffer material into account
Approaches based on taking weighted average strength
See Bowles, Das or other text B
soft clay
trending
stronger 2B
stiff clay or
dense sand

37. 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
dense sand
soft clay

38. 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