Soil Mechanics Solutions To Simple Problems

CIVIL 2 SOLUTIONS TO SOIL MECHANICS ASSIGNMENT CIVIL II 1.

W.T SAND Υ =1.92Mg/m3

0.0

-3.0 CLAY ΥS = 2.0Mg/m3

-7.5 Medium Sand

(a) Variation of Total Vertical pressure (i)

At -3.0m

σ=Υ

SX g X z = 1.92 X 10 X 3

SOIL MECHANICS GROUP ASSIGNMENT [Type the abstract of the document here. The abstract is typically a short summary of the contents of the document. Type the abstract of the document here. The abstract is typically a short summary of the contents of the document.]

= 57.6KPa At -7.5m

(ii)

σ=Υ

Sx g x z = (2.0 X 10 X 4.5)+57.6 = 147.6KPa

(b) Effective Vertical Pressure σ'= σ-µ Where µ=neutral stress or up thrust µ=ΥwXgXz At -3.0m, µ=1X3.0X10=30Kpa Therefore σ'=57.6-30=27.6KPa At -7.3m, µ=(1X4.5X10)+30=75Kpa Therefore σ'=147.6-(75)=72.6Kpa Table of variation of pressures with depth Depth(m) σ(KPa) µ(KPa) 0.0 3.0 7.5

0.0 57.6 147.6

0.0 30 75

σ'(KPa) 0.0 27.6 72.6

Diagram showing Total and effective vertical pressure Total vertical Pressure Effective vertical Pressure 0.0

0.0 57.6Kpa

27.6Kpa

-3.0

72.6Kpa

-7.5

147.6 Kpa k After the water table is lowered by 2.0m At -2.0m, µ=0 therefore σ'= σ=2.0X1.92X10=38.4KPa At -3.0m, µ=10X1=10Kpa, therefore σ'=(1.92X10X1+38.4)-10=47.6Kpa At-7.5m, µ=10X5.5=55KPa,therefore σ'=(2X10X4.5+57.6)-55=92.6KPa Table of pressure variation after water table is lowered by 2m Depth(m) σ(KPa) µ(KPa) σ'(KPa) 0.0 2.0 3.0

0.0 38.4 57.6

0.0 0.0 10

0.0 38.4 47.6

7.5

147.6

2.

WT GRAVEL Υ =2.0Mg/m3

55

92.6

-2.75 SATURATED CLAY ΥSC = 1.84Mg/m3 IMPERMEABLE ROCK

-17.75

Total Vertical Pressure (i)

At -2.75m Xz = 2 X 10 X 2.75 = 55Kpa

(ii)

At-17.75m = 1.84 X 10 X 15+55 = 331KPa

Up thrust Where µ=neutral stress or up thrust µ=ΥwXgXz At -2.75m µ=1X10X2.75 =27.5Kpa At -17.75m µ=1X10X15+27.5 =177.5Kpa

Effective vertical pressure σ'= σ-µ At -2.75m =55-27.5 =27.5Kpa

σ = ΥX g σ = ΥX g X z

At -17.75m =331-177.5 =153.5Kpa Table of pressure variation Depth(m)

σ(KPa)

µ(KPa)

σ'(KPa)

0.0 2.75 17.75

0.0 55 331

0.0 27.5 177.5

0.0 27.5 153.5

Case II GRAVEL Υ =1.57Mg/m3 WT -2.75 SATURATED CLAY ΥSC = 1.84Mg/m3 IMPERMEABLE ROCK

-17.75

Total Vertical pressure (i)

At -2.75m

σ = ΥX g X z = 1.57 X 10 X 2.75 = 43.175Kpa (ii)

At -17.75m = 1.84 X 10 X 15+43.175 = 319.175KPa

Upthrust Where µ=neutral stress or up thrust µ=ΥwXgXz At -2.75m µ=0X10X2.75 =0Kpa At -17.75m µ=1X10X15 =150Kpa

Effective vertical pressure σ'= σ-µ

σ = ΥX g X z

At -2.75m =43.175Kpa At -17.75m =319.175-150 =169.175Kpa

Immediate changes Pore water pressure =150-177.5 =-27.5Kpa A drop in power water pressure is observed from case 1 to case 2 Effective vertical pressure =169.175-153.5 =15.65Kpa A rise in effective vertical pressure is observed from incidence 1 to incidence 2 Table to summarize changes Depth(m)

σ(KPa)

µ(KPa)

σ'(KPa)

0.0 2.75 17.75

0.0 43.175 319.175

0.0 0.0 150

0.0 43.175 169.175