Settlement prediction of spread foundations

(i.e. strip/pad footings and rafts)

Accuracy of settlement predictions 4 footing tests (0.7m < B < 1.5m) in Perth Sand Prediction competition organised to estimate footing settlement of each footing at loads of 100 & 180kN; over 50 predictions received from around the world

Predictions of footing settlement at 100 kN and 180 kN

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19 9 20 18 10 2 1 13 14 4 11 26 16 15 5 23 24 28 6 21 27 3 17 25 12 8 22 7

Participiant number

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Score of 100:average of 8 predictions are within 20%Score of 60: average of 8 predictions within 50%

Score of 20: average of 8 predictions within a factor of 2Negative score: average of 8 predictions off by more than a factor of 4

Competition winner could only predict settlement to within a factor of about 2 !

Approaches to settlement prediction

1. Linear elastic half space (homogeneous)

w = π/4 qB (1-υ2)/E = π/8 (1-υ)qB/G (rigid footing) w0 = qB (1-υ2)/E = (1-υ)qB/2G (centre of flexible footing) wc= [(1-υ)qB/2G] ×I where I =f (L/B) (corner of flexible footing) 2. General linear elastic

• Use above formula and select E or G from standard correlations at depth = ZI/2

where ZI (m) = [B(m)]0.75

• Burland & Burbridge SPT N method and Schmertmann CPT qc method • E/G linearly increasing with depth e.g. G= Go + mz w0 = [qB/4G0]× I (centre of flexible footing, where I = f (2Go/mB, υ) • Finite thickness of compressible soil, assume settlement distribution from

maximum at footing level to 10% of this maximum at 2zI beneath footing and zero at 4zI beneath footing, where zI (m) = [B(m)]0.75

• (Computer based) derivation of settlements in non-homogeneous elastic material using Boussinesq’s equations

3. Approximate non-linear analysis

• Measurement of (non-linear) stiffness in triaxial tests • 1-D settlement prediction • Approximate means of applying non-linear triaxial stiffness data to settlement

prediction • Approximate non-linear settlement prediction in cohesionless soils • Stress-level and strain level dependent soil stiffness

4. Full non-linear analysis

• Use Finite Element (or Finite Difference) method and realistic soil constitutive model, incorporating stress and strain level dependence of stiffness, anisotropy etc.

Settlement of shallow foundations For 'typical' loading beneath shallow foundations (of width B and under bearing pressure =q): Settlement (rigid foundation) = (π/4) qB (1-υ2)/E’v

Settlement (centre of flexible foundation) = 1.1 qB (1-υ2)/E’v Where E’v is the equivalent drained Youngs modulus (Eeq) and: υ =0.2 E’v ≈ 5 qc or 2500 N60 (kPa) (in natural cohesionless soils) ≈ 750 N60 (kPa) or 2 qc (normally consolidated, unaged soils) ≈ 300 sutc (in high OCR clays) ≈ 150 sutc (in low OCR clays)

Alternatively the following plot from Baldi et al. (1988) is in popular use for sands:

E/qc =f (qc, stress history, stress level)

Using Burland & Burbidge SPT approach for sands and gravels Empirical correlation based on backanalyses of over 200 case histories (Note non- proportional relationship between the applied stress and the foundation width)1. w = fs fL [qn – (2/3) σ’vy]B0.7 Ic

• w (mm) = settlement • qn = net foundation pressure (=q – σv) in kPa • σ’vy = preconsolidation stress (kPa) • Ic = 1.7/N1.4 • N is the average SPT N value to depth = zI • zI (m) = [B(m)]0.75 • fs = shape correction factor ≈ [1.25 L/B/(L/B+0.25)]2 • fl=creep correction factor (=1.5 for static loads and 2.5 for fluctuating loads)

Schmertmann (1970, 1978) method

Schmertmann recommends E=2 qc The creep correction factor (C2) is often used in conjunction with other settlement prediction methods

1 This arises because of the inherent non-linear stress-strain characteristic of soils

1.0)(log12

2

02yearstCwherez

EIqCw z

B

n +=ΔΣ==>

Iz 0.6

2B

B/2

For shallow square footing

Example: Building with 8m column grid with SWL column load of 800 kN

To ensure satisfactory SLS, assume maximum tilt between column locations =1/400 and that adjacent column founded on ‘hardspot’ i.e. footing for this column does not settle.

Maximum allowable settlement = column spacing/400 = 20mm

s= 0.02= (π/4) B {Fcol/B2] (1 – 0.22)/15000 and Bmin=2m Check ULS for 2m square footing: q = 800/(22) = 200 kPa Typical φg=0.5 and load factor (γf) =1.35 Require 1.35 × 800 /22 =270 kPa to be less than

φg Nc sc dc su = 0.5 ×(2+π) × 1.2 ×1 × 50 = 154 kPa not OK Need to increase B to provide adequate FOS for ULS. The footing’s factor of safety will increase with time as the clay begins to drain under slow application of applied load from structure. Exercise: Repeat the above example assuming the soil stratum is sand. Compare the settlement given by the equation for a rigid punch with the settlement predicted by Burland & Burbridge.

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SPT N

Dep

th (m

)

Soil stratum is clay Need to found footings at minimum depth of ≈0.5 to 0.75m to avoid seasonal effects Assume footing formation level at 1m depth Average N value ≈ 10 sutc (kPa) ≈ 5 x N =50 kPa (high OCR) E’v ≈ 300 sutc (equivalent linear modulus)

E’v = 15000 kPa Assume ν=0.2

Settlements in non-homogeneous elastic soil and beneath flexible foundations

Example: Determine the settlement beneath the centre and corner of a flexible raft

(20m × 10m) under a building load of 20 MN on a clay with sutc=50 kPa underlain by

intact rock at a depth of 10m. Estimate the settlement at a lateral distance of 30m from

the centre of the foundation.

10m 10m

5m

5m

wc = (1-υ) qB I/2G Flexible raft q = 20000/ (20 × 10) = 100 kPa Eeq = 250sutc = 12500 kPa => Geq= 12500/[2(1+υ)] =5200 kPa for υ=0.2 For small (10m × 5m) rectangular loaded area, L/B=2, I=0.766 (from table) Settlement at centre of raft in half space (w0) = 4 (rectangles)× [ (1-0.2) × 100 × 5 × 0.766/ (2×5200)] =0.118m = 118mm =wmax Settlement at corner of large rectangle in elastic half space (L/B=2); I also =0.766 = 1 (rectangle) × [ (1-0.2) × 100 × 10 × 0.766/ (2×5200)] =0.059m = 59mm =wcmax zI beneath foundation = 100.75 = 5.6m w =0.1 × 118mm at depth of 2 × 5.6= 11.2m w below 10m depth in elastic half-space

= [0.1 + (1.2/11.2)×0.9] = 0.2 wmax w at centre of raft = (1-0.2) × wmax = 90 mm w at corner of raft = (1-0.2) × wcmax = 47mm Approx. settlement (in linear elastic soil) at distance (s) of 30m from centre: [(1-υ)Q/(2Gπ)]/s = [(1-0.2) × 20000/ (2 x 5200 x π)]/30 =0.016m = 16mm

Elastic Settlement prediction using Boussinesq’s equations

Estimate equivalent linear elastic vertical and horizontal moduli for soil horizons beneath

a uniformly loaded area. In the absence of additional information, assume E’v=E’h and

calculate the vertical strain at the centre of each horizon from:

Δεz = [Δσ’v /E’v ] – [ 2νΔσ’h/ E’v] (1)

where ∆σ’v and ∆σ’h are determined at the mid-height of each soil horizon using

Boussinesq’s equations. The settlement at the foundation level is then calculated by

summing the settlements of each soil horizon (=∆εz ∆z) from a point below the

foundation where settlements are assumed to be zero up to the foundation level i.e.

w = Σ Δεz Δz (2)

The settlement calculated beneath the centreline of the foundation may be reduced by (π/4)

to give the settlement of a rigid pad foundation. Instead of using Boussinesq equations (via

computer programs such as OASYS VDISP), charts such as those given on page 22 may be

used to estimate stress distributions. A Poissons ratio (υ) of 0.2 is usually (conservatively)

employed using equation (1).

The approach is not recommended for calculating distributions of settlements at some

distance from the loaded area because the assumption of linear elasticity often leads to

under-prediction of settlement gradients and over-prediction of settlements.

2

2 Classification system is based on ease of repair of plaster and brickwork or masonry. The Burland & Wroth (1974)

classification system is commonly used with categories 0,1,2,3,4 and 5. For Category 0 (negligible damage), hairline cracks (less

than 0.1mm) are barely visible. For Category 2 (slight damage), re-decoration is probably required, crack widths are up to 5mm

wide but are easily filled. For Category 4 (severe), windows and door frames are distorted, walls are leaning and/or bulging,

crack widths are up to 25mm. Most damage occurs because cladding and finishes experience tensile strains. At strains in excess

of 0.05% to 0.1% (measured over a gauge length of >1m), visible cracking occurs. At ‘damage categories’ 0, 2 and 4, tensile

strains are typically ≈0.025%, 0.1% and in excess of 0.3% respectively. The two expressions for Δ/Lε given above are typical of

a range of predictions summarised in Burland et al. (2004)

Bending: ∆w/(Lεlim) ≈ L/6H + 2H/3L (neutral axis at H/2)

Shear: ∆w/(Lεlim) ≈ 1 + L2/4H2 (neutral axis at H/2)

εlim <0.05% => no damage εlim>0.3% => severe damage

Tunnelling induced movements

For bending in load bearing wall:

Δh/(Lhεlim) ≈ 0.4

Δs/(Lsεlim) ≈ 1.5-3

=> Hogging is critical

EXAMPLE: The feasibility of employing a raft foundation for a 20 storey building

and a plan area of 100m × 100m is to be investigated. The equivalent design soil

shear modulus (with υ=0.3) was assessed as being: G= 29 +2z (MPa), where z is

depth below the raft. Assume a design storey loading of 12 kPa, a raft thickness,

Youngs modulus and Poissons ratio of 1.5m, 30 kN/mm2 and 0.3 respectively and an

allowable maximum tensile strain of 0.05%.

Using relationship ∆w/(Lεmax) = L/6H + 2H/3L and estimating H=80m=>

∆w (allowable) = [100 × (0.05/100) × (100/480 + 160/300)]m = 37 mm

Gross bearing pressure (q) at underside of raft = 20 × 12 + 1.5 ×24 = 276 kPa

Raft zone of influence (zI) ≈ 1000.75 = 32m

Average G ≈ 29 + 2 (zI/2) => G = 29 + 2 × 32/2 = 61 MPa

Average E = 2 × 48000 × (1+υ) = 160 MPa

For flexible raft: Select I values for rectangular areas on homogeneous soil

• At raft centre: w = 4 × [(1-0.3)(276 ×50×0.561) / (2 × 61000)]m = 178mm3

• Mid-side : w = 2 × [(1-0.3)(276×50×0.766)/(2 × 61000)]m= 121mm

• Corner of raft: w = 1 ×[(1-0.3)(276×100×0.561)/(2×61000)]m = 89mm

For rigid raft : w =π/4 qB (1-υ2)/E = π/4 × 276 × 100 ×(1-0.32)/160000 = 123mm

Differential along central axis = 178-121= 57mm > 37mm (allowable => need raft

stiffness to reduce differential – subject to Structural Engineer’s specification)

Differential along side walls (centre to corner = 121-89 =32mm < 37mm allowed)4

3 Note: this approach implies w is 12% higher than the more commonly adopted expression, w=qB(1-υ2)/E 4 Although, elastic solutions often under-estimate settlements on the periphery of actual foundations

To limit tensile strains within central area of raft, require ∆w/wrigid < 37/123 = 0.30

From chart, require Krs >0.05

Krs for 1.5m thick raft (with Econcrete = 30,000 MPa)

= 5.57 × (30000/160) × (100/100)0.5 × (1.5/100)3 = 0.0035 < 0.05

For Krs =0.05 = 5.57 × (30000/160) × (100/100)0.5 × (t/100)3

=> Raft thickness, t, required = 3.6m

Note that the raft flexibility is relatively high because L=100m

For Krs=0.05 M/qL2 = 0.0025 where M is the bending moment per unit length

q=276 kPa, L=100m => M =0.0025 × 276 × 1002 = 6900 kNm/m

Calculate steel requirement from M/bd2 i.e. Asteel = f (M/bd2, fc)

M/bd2 = 6900/(1.0 × 3.62) = 532 kN/m2 = 0.532 N/mm2

=> low/nominal steel required

Usual to employ soil-structure interaction (FE) analysis to optimise design

Approximate Non-linear settlement prediction methods

• Measurement of (non-linear) stiffness in triaxial tests • 1-D settlement prediction (data provided by Oedometer tests) • Approximate means of applying non-linear triaxial stiffness data to

settlement prediction • Approximate non-linear settlement prediction in cohesionless soils • Stress-level and strain level dependent soil stiffness

Generally unaccountable errors remains due to (a) difficulties in trimming the sample precisely, (b) play in connection between load cell and sample top cap and (c) bedding down ends of sample due to local surface irregularities

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Axial strain (%)

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r stre

ss (k

Pa)

External axial strain measurement

Internal axial strain measurement

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Axial strain (%)

Sec

ant Y

oung

s m

odul

us (M

Pa)

Internal strain measurement

External strain measurement

Effect of local vs. external strain measurement on inferred stiffness

Volume gauge

Used for triaxial testing • Measures deviator stress • Eliminates piston friction • Very insensitive to transverse forces

Example5: Estimate the mean settlement of a 50m × 50m flexible raft with a bearing pressure of 80 kPa founded on the surface of a 15m thick layer of soft clay overlying rock. The available properties for the clay are as follows. Assume the water table is at ground level. w = 45%, LL=50%, Pl=20% σ’vy =100 kPa throughout the layer Estimates: Assumption of 1-D is valid

e ≈2.7w = 1.22 and γb=γw (2.7+1.22)/(1+1.22) = 17 kN/m3

Cc* = 0.256 eL -0.04 = 0.256×2.7×0.5-0.04 =0.35

Cc ≈1.3Cc*=0.45, Cs≈0.2Cc

*=0.07, Cα =0.05 × Cc =0.0225 Clay when consolidated will mostly be at OCR=1, take cv =2m2/year Time for consolidation =7.52/2=28 years (assuming 2-way drainage)

Rock (assumed rigid & permeable) Divide 15m layer into 3 5m thick sublayers Sublayer No. z (at centre of sublayer) σ’vi (kPa) σ’vf(kPa) σ’vy (kPa)

1 2.5m 17.5 97.5 100 2 7.5m 52.5 132.5 100 3 12.5m 87.5 167.5 100

Consolidation settlement (mm) = [(0.07/2.22) log(97.5/17.5)]5000

+ [(0.07/2.22)log(100/52.5) +(0.45/2.22) log(132.5/100)]5000 + [(0.07/2.22)log(100/87.5) +(0.45/2.22) log(167.5/100)]5000 ≈525mm

Creep over 100 year life of structure = (0.0225/2.22) ×15000×log(100/28) ≈ 84 mm Total settlement ≈610 mm

5 c.f. correlations presented earlier in the unit

Additional methods for predicting settlement of shallow foundations Non-linear method for footings on clays: using triaxial (local strain) data

Derive equivalent modulus from high quality samples tested appropriately under triaxial conditions. For foundation of width, B, and settlement w, the equivalent modulus is approximately the same as secant modulus obtained in triaxial tests when the triaxial axial strain (εv) is 2.5 times less than w/B. An iterative procedure is required:

1. Estimate w/B 2. Determine Eeq (E’ or Eu) at εv =0.4(w/B) from triaxial stiffness data 3. Calculate w from w=(π/4)qB (1-υ2)/Eeq 4. Calculate new w/B and return to step 2.

(see Tutorial example) Non-linear method for footings sands using in-situ test data Use is still made of elastic equation s/B = π/4 q (1-ν2)/ Eeq, but Eeq is allowed to reduce as strain level (and s/B) increase. In addition a more realistic relationship between Eeq and qc is employed.

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0.01 0.1 1 10s/B (%)

E eq/q

c

B=3mB=2.5mB=3mB=1mB=1.5mB=1.5mB=1mB=1mB=0.67m

Texas: Closed symbolsShenton Park: Open symbols

Backfigured equivalent moduli (Eeq) from 9 test footings at two sites => Eeq reduces with s/B At given s/B, Eeq is not proportional to qc

A~11 and C~250 in Perth (normally consolidated) dune sand A~19 and C~800 in aged alluvial Texas sand

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400

0.01 0.1 1 10s/B (%)

E eq/

(qc0.

25 σ

' v00.

5 pa0.

25)

B=3mB=2.5mB=3mB=1mB=1.5mB=1.5mB=1mB=1m SPB=0.67m SP

Shenton Park: Open symbols

Texas: Closed symbols

BsAE

pqE eq

avc

eq

/' 25.05.025.0 =≈ξσ

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛≈=>

acv pqqq

CBs

0'1

σ

2

2 )1(4 ⎟⎟

⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

−=

νπACwhere

σ'v0 and qc are calculated at a depth below the footing of ZI/2, where ZI (m)= [B(m)]0.75

Using the DMT to assess likely value of A and C (These values depend on age, OCR of deposit etc e.g. see Baldi plot of E/qc)

Expansion of membrane to s/D=1.8% Lift-off pressure =p0 Pressure at s/D=1.8% = p1 KD = (p0-u0)/σ’v0 From elasticity ED=34.7 (p1-p0) Eeq for footing varies with ( ED/KD

0.5) =>at s/B=1.8%, ratio of A values = ratio of ED/KD

0.5 values e.g At Shenton Park sand, A =11 and average ED/KD

0.5 value=4.5, A value at another site with DMT parameters of ED1 and KD1 = 11×(ED1/KD1

0.5)/4.5

r = (ED1/KD10.5)/ (EDSP/KDSP

0.5), where EDSP and KDSP are DMT parameters at Shenton Park

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400

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s/B (%)

E eq/

(qc0.

25 σ

' v00.

5 pa0.

25) =

Eeq

/ ξ

Shenton Park

Texas

1.8%

75

75r

Creep settlement of footings in sand The settlement of footings increases with time and a linear variation of settlement with the logarithm of time is usually observed i.e. where sc is the settlement due to creep and m is a creep coefficient which increases with the levels of strength mobilisation. For footings on normally consolidated (Perth) sand, the following expression provided a good fit to footing settlement data (and also to creep occurring during stress hold tests in pressuremeter tests) The value of m for footings on overconsolidated sand may be expected to be less than one third of the value given by the above expression, while m for footings on normally consolidated sands subjected to drained cyclic loads is likely to be double that given by the same expression. The total settlement is the sum of the ‘immediate’ and creep settlement: Where m1=0.02 for typical Perth dune sand

hourstwithqqqqm reff

f

27.0/02.0~2

=−<⎟⎟⎠

⎞⎜⎜⎝

⎛

tm

dtBsd

ttm

Bs

nc

ref

c s ===>⎟⎟⎠

⎞⎜⎜⎝

⎛=

•)/(ln

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟

⎟⎠

⎞⎜⎜⎝

⎛≈

)24/2()(

ln '

12

10

2 daystqqm

pqq

CBs

facvσ

Using stress and strain level dependent Youngs Modulus

• E0/pa = AE F(e) (σ’v/pa)0.5

• F(e) = (2.17 –e)2/(1+e)

E0

• Esec = _______________ for ε > εel 1 + (ε-εel)n/(εr-εel)n • εr = Cr (σ’h/pa)0.5

• Δεz = [Δσ’v /E’v ] – [ 2νΔσ’h/ E’v]

• s = Σ Δεz Δz

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Axial strain (%)

E' v/

E 0

n=1 εr=0.1%

n=0.7 εr=0.01%

εel=0.0008%

EXAMPLE: The CPT end resistances (qc) and shear wave velocities (Vs) measured in a SCPT at a sand site in Perth are shown below. The K0 value of the sand deposit, which has an in situ mean water content of 3%, may be assumed =1.0 and the void ratio limits are 0.8 and 0.4. A 10m × 10m raft is to be founded at ground level. (i) Assess the mean density and relative density of the sand. (ii) Derive a relationship between the very small strain elastic modulus (E0) of the

material and the vertical effective stress. (iii) Estimate the settlement response at the centre of the raft at a maximum bearing

pressure of 300 kPa assuming an elastic stress distribution beneath the raft. The secant Youngs modulus (Esec) for the sand may be obtained from the following expression derived from triaxial data:

E0/ Esec = 1 + [(ε-εel)/(εr-εel)]0.6

where ε is the vertical strain, εel (the linear elastic limit) is 0.0001% and εr (%) =0.01 (σ’h/pa)0.5; εr is the strain at which Esec/Eo=½ and σ’h is the horizontal effective stress.

Assume Poissons ratio values of 0.1 and 0.4 for very small strains and larger strains respectively.

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Shear wave velocity (m/s)

Dep

th (m

)

(Rigid) siltstone

Perth

san

d

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CPT qc (MPa)

Dep

th (m

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∆σz = αz qapp ∆σh = αr qapp ∆p= αp qapp ∆q = αq qapp qapp = applied stress z = depth below centreline R= Equivalent foundation radius

emax emin Dr e w Sr ρb/ρw γb

(kN/m3) 0.8 0.4 0.5 0.6 0.03 0.1325 1.706 16.73525

At small strains

vo =0.1

Dr qc

(kPa) z

(m) vs (m/s) Go (MPa) Eo/pa p'=σ'v (σ'v/pa)0.5 F(e) AE D 0 0 (K0=1) 0.47 7000 2 180 55 1216 33.5 0.578537 1.54056 1364.3 21018 0.47 10000 4 200 68 1501 66.9 0.818175 1.54056 1191 18348 0.53 14500 6 220 83 1816 100.4 1.002055 1.54056 1176.7 18128 0.57 19000 8 240 98 2162 133.9 1.157074 1.54056 1212.7 18683 0.51 18000 10 250 107 2346 167.4 1.293648 1.54056 1177 18132 0.50 19000 12 280 134 2942 200.8 1.41712 1.54056 1347.8 20763 0.50 20000 13.5 300 154 3378 225.9 1.503083 1.54056 1458.7 22472

Average Dr=0.5 Average 1275.5 19649 Eo/pa= 1275*F(e)*(σ'v/pa)0.5 = 2000(σ'v/pa)0.5 for Dr=50% Eo (kPa) 19500 σ'v0.5 for Dr=50% with σ'v in kPa

Parts (i) and (ii)

Part (iii): Calculations shown for applied stress of 300 kPa B 10

a 5.6419(equivalent radius)

γb 16.7352 Κο 1 D 19500 εel (%) 1.00E-04 Cr 1.00E-02 v 0.4 n 0.6 Stress 300 kPa (1) Circ ref

z-middle z/a αz αh Δz

σ'vi (kPa)

σ'hi (kPa) Δσ'v Δσ'h σ'vm σ'hm

Eo (kPa) εr (%)

Δσ'v - 2vΔσ'h ε

Esec (kPa)

Set =column (1) s (mm)

0.6 0.1 1 0.8 1.13 15 15 300 240 165 135 250817 0.012 108 4.10E-03 26365 0.00409635 5

1.7 0.3 1 0.6 1.13 34 34 300 180 184 124 264745 0.011 156 8.65E-03 18028 0.00865339 10

3.4 0.6 0.9 0.3 2.26 63 63 270 90 198 108 274147 0.010 198 1.52E-02 13027 0.01519896 34

5.6 1 0.66 0.15 2.26 100 100 198 45 199 123 275370 0.011 162 8.68E-03 18653 0.00868488 20

7.9 1.4 0.5 0.05 2.26 138 138 150 15 213 146 284717 0.012 138 5.07E-03 27195 0.00507445 11

11.3 2 0.28 0 4.51 195 195 84 0 237 195 300095 0.014 84 1.40E-03 59951 0.00140114 6 Total 86mm