Foundation Settlements -...
Transcript of Foundation Settlements -...
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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