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CE-5133 (3 Credit Hours)

Geotechnical and Foundation Engineering

Foundation Settlements

Instructor:

Dr Irshad Ahmad

Lecture on 15-4-14

Department of Civil Engineering

University of Engineering and Technology, Peshawar

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Stress strain modulus Es

Stresses in Soil Mass from Elastic Theory

Contact Pressure Distribution

Contents

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Settlement Problems (Bowels P-284)

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Settlement Problem (Bowels P-284)

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Determination of Stress Strain Modulus Es (B P-124)

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Initial tangent and Secant Stress Strain Modulus Es

(B P124)

E is also determined as secant modulus between origin and 1/3 of the peak stress, or over the actual stress range in the particular problem

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Determination of Stress Strain Modulus Es (B P-313)

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Determination of Stress Strain Modulus Es (B P-313)

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Determination of Stress Strain Modulus Es (B P-313)

U=Unconsolidated Undrained Triaxial Test CU= Consolidated Undrained Triaxial Test CKoU=Consolidated to Ko and tested undrained i.e Consolidated with some vertical pressure and with lateral pressure set to an estimate of field value of Kov

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Determination Es (B P-102)

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In situ Tests: Stress Strain Modulus Es (B P-314)

Because of the effects of sampling disturbance, it is preferable to determine E (or G) from the results of in-situ tests.

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B P-316

qc is point resistance In CPT

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Bowels P123

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Pressuremeter Test

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Dilatometer test (DMT)

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Finding Secant Modulus Es (B P-324)

=35 Ko=1-sin=0.426 Overburden pressures: v1 at No.1 = 34.6 kPa Overburden pressures: v2 at No.2 = 80.6 kPa Lateral earthpressure 3 at No.1 = Ko v1 = 14.7 kPa (use 20 kPa) 3 at No.2 = Ko v2 = (use 40 kPa) Run triaxial tests under 20 and 40 kPa of cell pressure – i.e. consolidating under Ko condition

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Determination of Stress Strain Modulus Es (B P-324)

(1-3) in test corresponding to the site is q: q = qv – qh = qv – Ko qv q = qv (1-Ko) = P/A (1-Ko) = 2100/9 (1-0.426) = 133.9 kPa So (1-3) = 133.9 kPa Secant Modulus (curve-1): 1=7x10-3 for (1-3) = 133.9 kPa Secant modulus Es = 133.9 / 7e-3 = 19130 kPa

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Shear Modulus (G) and Poisson’s ratio or

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Poisson’s ratio (Bowels P123)

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Stresses from Elastic Theory

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Variation of vertical stress due to point load

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Stresses due to line load

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Strip Area carrying a uniform pressure

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Depth to 0.1q for Strip Area with q

z = q/ { + sin() cos(+2)}

= -/2

0.1q = q/ { + sin() cos(0)}

0.314 = {/57.3 + sin()}

=9

Tan(/2) = B/2z

Z = 6.5B

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Contours of Equal Vertical Stress under strip area

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EXAMPLE- Soil Mechanics by Craig

Answers (kN/m2) 30.6, 12.2, 129.6,15.6

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Circular area- uniform pressure

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Rectangular area carrying uniform pressure

(Bowels 295)

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Rectangular area carrying uniform pressure

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Stress under rectangular area

Answers (kN/m2) 44

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Slope 2:1 Method (Bowels 286)

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Foundation Settlements (Bowels 286)

=Q/(B+z1)(B+z2_)

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Problem (Bowels P331)

Find average stress increase in 10 ft thick clay layer using slope 2:1

method

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Problem (Bowels P331)

Find average stress increase in 10 ft thick clay layer using slope 2:1

method

Z1=6ft, z2= 16 ft, H=10ft, Q=375 kips, B=8 ft

qv, average increase in stress = -375/10 [ 1/(8+16) – 1/(8+6) ]

= -37.5 (-0.02976) = 1.11 ksf

=Q/(B+z1)(B+z2_)

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Problem (Bowels P331)

Find average stress increase in 10 ft thick clay layer using Boussinesq

equation

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Problem (Bowels P331)

Divide the 10ft clay layer into four sub-layers each 2.5feet thick

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Problem (Bowels P331)

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Problem (Bowels P331)

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Assignment (Bowels P331)

Resolve the problem assuming the footing is circular 9 feet

diameter.

9 feet dia is chosen to get

same contact stress as for

the 8x8 feet rectangular

footing.

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Assignment (Bowels P331)

Apply Simpson’s Rule: z(avg)= 2.5/10 [ (2.877+0.63)/2 + 1.827 + 1.22 + 0.86 ] = 1.415 ksf

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Contact Pressure: Behavior of Clay and Sand

Rigid slab: Contact pressure: Non uniform Deflection: Uniform

Flexible slab: Contact pressure: Uniform Deflection: Non uniform

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Saint-Venants Principle