University of Melbourne
Department of Infrastructure Engineering
CVEN90027 Geotechnical Application 2015

Assignment 3
Foundation Design Component

Kaijie Zheng
745870

Check for bearing capacity
For stiff clay layer qapplied,net=120-P0 P0=19×5=95kPa
Thus, qapplied,net=120-95=25kPa
Use Terzaghi equation to analysis.
For clay, φu=0°, thus only the first term of Terzaghi equation is required. qu net=(cNc)1η0
Use Skempton’s approximation for Nc, with B=28m, L=28m and D=5m.
Nc=51+0.2BL1+0.2DB=5×1.2×1+0.2×528=6.2

Use a quarter of area, thus B'=14m, L'=14m Z/B'=10/14=0.71, which gives η0=0.21.
Thus, qu net=120×6.2×10.21×4=885.7kPa
On clay layer, FS=885.725=35.4.
For medium dense silty sand layer
Use Terzaghi equation of square foundation qu net=1.2cNc+P0'Nq-1+0.4BγeffNγ
P0'=95kPa. For sand, c=0, thus the first term could be eliminated. For φ'=35°, Nq=Nγ=40. γeff=20-10=10kN/m3.
Thus, qu net=95×39+0.4×28×10×40=8185kPa.
On sand layer, FS=818525=327.4
For dense gravel layer

It follows that the stiff silty clay layer is more critical.

Settlement calculation
Immediate settlement for clay layer
Consider 14 of area, B'=14m, L'=14m, H=10m.
L'B'=1, HB'=0.71.
From chart of influence values, F1=0.1, F2=0.075.
Iρ=F1+1-2v1-vF2=0.1+1-2×0.51-0.5×0.075=0.1
ρi=qB(1-v2)EIρ q=120kPa, Eu=45MPa, vu=0.5
Thus, ρi=4×120×14×1-0.2545×0.1=11.2mm
Consolidation settlement in clay layer
Initial effective stress at midpoint of the layer: σo'=10×19=190kPa Z(m) | ∆H(m) | σo'(kPa) | Z/B | Iσ | ∆σ'(kPa) | σf'(kPa) | σave'(kPa) | mv(MPa-1) | ρ(mm) | 5 | 10 | 190 | 0.18 | 0.95 | 23.75 | 213.75 | 201.875 | 0.05 | 11.875 |
∆σ'=0.95×120-95=23.75kPa
σf'=190+23.75=213.75kPa σave'=190+213.752=201.875kPa mv=0.05 from Table2 provided ρc=mv∆H∆σ'=0.05×10×23.75=11.875mm Total settlement in clay: ρT=DF×RF×(ρi+μρc) Rigid factor RF=0.8, Depth factor DF=0.98 as Z/B=0.18, μ=0.5
Thus, ρT=0.98×0.8×11.2+0.5×11.875=13.44mm.
Immediate settlement in sand layer
Use superposition of to analyse.

For 0-20m
Use 14 of area, B'=14m, L'=14m, H=20m
L'B'=1, HB'=1.43, v'=0.3, E'=80MPa
Thus, F1=0.24, F2=0.075
Iρ=F1+1-2v1-vF2=0.24+1-2×0.31-0.3×0.075=0.28

ρi=4×120×14×1-0.0980×0.28=21.4mm
For 0-10m
H=10m, HB'=0.71
From chart of influence values, F1=0.1, F2=0.075.
Iρ=F1+1-2v1-vF2=0.1+1-2×0.31-0.3×0.075=0.14
ρi=4×120×14×1-0.0980×0.14=10.7mm
21.4-10.7=10.7mm
Thus, the immediate settlement in sand is 10.7mm.
Immediate settlement in Gravel layer
For gravel to infinity, settlement is given by ρ=qB(1-v2)EIρ v'=0.25, E'=150MPa
Use 14 of area, for L'B'=1, Iρ=0.56 ρi=4×120×14×1-0.252150×0.56=23.52mm For gravel to 20m, F1=0.24, F2=0.075.
Iρ=F1+1-2v1-vF2=0.24+1-2×0.251-0.25×0.075=0.29
ρ0-20=4×120×14×1-0.252150×0.29=12.18mm ρ=23.52-12.18=11.34mm Thus, the immediate settlement in gravel layer is 11.34mm.
The maximum differential settlement calculation
Settlement in sand and gravel layers
10.7+11.34×0.8×0.98=17.3mm
Settlement on clay is 13.4mm.
Thus, there are roughly 56% sand and gravel, and 44% clay.
For a footing foundation, k=25000×0.44+15000×0.56=19400.
Total settlement ρT=11.2+10.7+11.34+0.5×11.875×0.98×0.8=30.7mm Thus, the differential settlement is:
∆ll=ρmaxk=30.719400=1630