# Geotechnical design

Páginas: 5 (1208 palavras) Publicado: 3 de abril de 2013

QUESTIONS 2
SOLUTIONS 2
Question 1 2
Question 2 5
Question 3 5
Question 4 6
Question 5 7
REFERENCES 9

QUESTIONS

1) Determine the total consolidation settlement of the clay layer after application of the maximum load or pressure from the tank. Assume the tank is founded at existing ground level.
2) Determine the overall settlement ofthe superficial deposits (above rockhead) after the application of the maximum pressure from the tank.
3) Determine the time required to reach 75% consolidation settlement.
4) Determine the bearing capacity of the soil at 4.0 m below ground level, assuming the tank will be founded at this level on a rigid 0.5 m thick reinforced concrete slab with minimum width of 6.0 m.
5) If the tank foundationsare to be piled, assume foundation level of 4.0 m below ground level with the maximum load of the tank and foundation slab (0.5 m thick reinforced concrete slab with dimensions matching the base dimensions of the tank) is equally distributed over thirty (30 no.) precast piles 0.30x0.30 cm2 cross-section, driven down though the alluvial sands/gravel to bear within the clay layer. Determine thepile length for FoS=2 on the ultimate bearing capacity.

SOLUTIONS
Question 1

Empty tank weight: 1554.13 KN
Tank dimensions:
- Width: B=7.87 m
- Length: L=9.15 m
- Height: h=5.00 m

Top of the clay (Z=10 m) Bottom of the clay (Z=20 m)
n=L⁄((Z)) 9.15⁄10 = 0.4575 9.15⁄20 = 0.2288
m=B⁄((Z)) 7.87⁄10 = 0.3935 7.87⁄20 = 0.1968
I_σ 0.066 0.020
σ_z=q× I_σ 70.63 × 0.066 × 4 = 18.65 70.63 × 0.020 × 4 = 4.8
Vertical Stress (kN/m2) 18.65 5.65

d_sand = 1780 Kg/m^3
γ_sand = (1780 x 9.81)/1000 = 17.46 KN/m^3
Before Construction
Top of clay Bottom of clayσ (KN/m^2) 17.46 × 10 = 174.62 174.62+(2.10×9.81×10) = 378.67
µ (KN/m^2) 9.81×6 = 58.86 9.81×16 = 156.96
σ' (KN/m^2) 174.62-58.86 = 115.76 378.67-156.96 = 221.71
Average (KN/m^2) ((115.76+221.71))⁄2 = 168.74

After construction
Top of clay Bottom of clay
σ (KN/m^2) 18.65+(17.46×10)=193.26 193.26+(2.08×9.81×10)+5.65=384.32
µ (KN/m^2) 9.81×6 = 58.86 9.81×16 = 156.96
σ' (KN/m^2)193.26-58.86 = 134.4 384.32-156.96 = 227.36
Average (KN/m^2) ((134.4+227.36))⁄2 = 180.88

Moisture content (w) = 31.8% = 0.318
Specific gravity (G_s )= 2.66
e_(0 )= w × G_s = 0.318 × 2.66 = 0.84588

Odometer Test Results
Sample Thickness (mm) 19.6 19.25 18.98 18.61 18.14 17.68 17.24 17.92
Change in Thickness (mm) 1.68 1.33 1.06 0.69 0.22 -0.24 -0.68 0
de0.158 0.128 0.103 0.068 0.022 -0.025 -0.073 0.000
e 1.0 0.973 0.949 0.914 0.868 0.821 0.773 0.846

Slope of e-p curve:
de/dp=((1.0-0.868))/((0-200))=-0.000679
m_v=-1/(1+e_0 )×de/dp==-1/(1+0.84588)∙(-0.000679)=3.6785×〖10〗^(-5)

Total Settlement of the clay layer:

C_s=m_v∙H∙∆σ=3.5755x〖10〗^(-5)×10×(180.88-168.74)=0.0447 m=44.7 mm

Settlement in the clay layer: 44.7 mm

Question 2Settlement= f_f×f_1×f_t×[q×B^0.7×I_c ]
f_s=(((1.25∙L)⁄B)/(L⁄B+0.25))^2=(((1.25∙9.15)⁄7.87)/(9.15⁄7.87+0.25))^2=1.6635
f_1=H/Z (2-H/Z)→Z=1,H=10∴H>Z
f_t=(1+R_3+R∙log t/3)=(1+0.7∙log1)=1
q = 70.63 KN/m^2
I_c = 6.5 × 〖10〗^(-2) (Graph)

Sand layer settlement = 1.6635 × 70.63 × 7.870.7 × 6.5 × 〖10〗^(-2) = 32.37 mm

Overall settlement = 44.7 + 32.37 = 77.07

Question 3

H = 5000 mmT=(0.403+0.567)/2=0.485 (From table)
Cv = 1.5 m^2/year
T=(T〖∙H〗^2)/C_v = (0.485+5^2)/1.5=8.1 years

Question 4

Bearing capacity for a foundation at 4.0 meters below the ground level.
Z = 4.0 m
B = 6.0 m
h_slab = 0.5 m
Soil type: Sand,loose to medium dense.
N_SPT = 10 blows
γ_sand= 17.46 kN/m3

From Meyerhof 1956, Foundation Engineering Handbook, a sand with N_SPT...

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