ECGD 4122 – Foundation Engineering Faculty of Applied Engineering and Urban Planning Civil Engineering Department 2nd Semester 2008/2009 ECGD 4122 – Foundation Engineering Lecture 4

Content Shear Strength Subsoil Exploration

Mohr-Coulomb Failure Criterion Total Stresses: Coulomb approximation for shear stress on the failure plane:

Mohr-Coulomb Failure Criterion

Shear Strength Measurement Based on Mohr’s circle criteria, two ways are possible to cause failure:

Shear Strength Measurement First: Increase the normal stress in one direction

Shear Strength Measurement Second: Apply shear directly

Shear Strength Measurement 3=0 1 Direct Shear Uniaxial Compression

Shear Strength Measurement 1 Unconfined compression test is used when  = 0 assumption is valid Triaxial compression is a more generalized version The soil sample is first compressed isotropically and then sheared by axial loading 3 Triaxial Compression

Direct Shear Test

Direct Shear Test

Direct Shear Test

Direct Shear Test

Direct Shear Test

Direct Shear Test

Direct Shear Test

Unconfined Compression Test 3 = 0 1 For clay soils Cylindrical specimen No confining stresses (i.e. 3 = 0) Axial stress = 1 Uniaxial Compression

Unconfined Compression Test 3=0 1 Uniaxial Compression

Unconfined Compression Test Data

Unconfined Compression Test Data

Unconfined Compression Test Data

Unconfined Compression Test 3=0 1 Uniaxial Compression  Max. shear plane Horiz. plane  1

Unconfined Compression Test

Triaxial Compression Test

Triaxial Compression Test

Triaxial Compression Tests Unconsolidated Undrained (UU-Test); Also called “Undrained” Test Consolidated Undrained Test (CU- Test) Consolidated Drained (CD-Test); Also called “Drained Test”

Consolidated Undrained Triaxial Test for Undisturbed Soils

Shear Strength in terms of Total Stress Shear Strength in terms of effective stress Shear strength in terms of total stress hydrostatic pore pressure

Shear Strength Total Stress -  = 0 condition Shear strength in terms of total stress For cohesive soils under saturated conditions,  = 0.

Mohr-Coulomb Failure Criterion Shear Strength,S  = 0 C Normal Stress, 

Subsoil Exploration Boring Sampling Standard Penetration Test (SPT) Vane Shear Test Cone Penetration Test (CPT) Observation of Water Table

Boring Boring Spacing

Boring ASTM Method

ASTM Method for Calculating Depth of Boring

Boring Methods Auger 3-5 m deep Hand-driven

Boring Methods Flight Auger 1-3 steps Tractor mounted

Boring Methods Rotary drilling Wash boring

Common Sampling Methods Standard Split Spoon The sampler is driven into the soil at the bottom of the borehole by means of hammer blows.

Sampling

Sampling

Sampling

Common Sampling Methods Standard Split Spoon The number of blows required for driving the sampler through three 152.4 mm (6”) intervals is recorded. The sum of the number of blows required for driving the last two 152.4 mm (6 in.) intervals is defined as the standard penetration number (N).

Standard Penetration Number

Standard Penetration Number

Vane Shear Test

Vane Shear Test

Vane Shear Test

Cone Penetration Test Measures: The Cone Resistance The Frictional Resistance

Cone Penetration Test

Cone Penetration Test

Cone Penetration Test

Assignment # 1 Determine the effective stress at point X for the two cases shown below. The drawing is not to scale. Take sat = 19 kN/m3.

Assignment # 1 Static condition: s = (9.81)(0.6) + (19)(1) = 24.89 kPa u = (0.6 + 1)(9.81) = 15.70 kPa s’ = s – u = 9.19 kPa Flow-down condition: i = h/L = 0.6/3.0 = 0.2 u = (0.6 + 1)(9.81) - (0.2)(1.0)(9.81) = 13.73 kPa s’ = s – u = 11.16 kPa

Assignment # 2 A soil deposit is composed of clay with Sat = 20 kN/m3. The GWT is at the ground surface. Calculate the shear strength on a horizontal plane at a depth of 10m, if c’ = 5 kPa and ’ = 30º.

Assignment # 2 t = c’ + s’tanf’ s’ = (20)(10) – (9.81)(10) = 101.9 kPa  t = 5 + (101.9)(tan30º) = 63.83 kPa

Assignment # 2 A drained direct shear test is performed on a normally consolidated silty soil. The soil sample is 75 mm in diameter and 25 mm in height. The vertical load is 883 N and the shear force at failure is 618 N. Calculate c and .

Assignment # 2 Drained test  t = c + stanf Normally consolidated soil  c = 0 t = c + stanf  t = stanf  f = tan-1(t/s) t = S/A = 618/[(37.510-3)2] t = 139.89 kPa s = N/A = 883/[(37.510-3)2] s = 199.87 kPa  f = tan-1(139.89/199.87)  35º

Quiz # 1 Determine the height of water level (H) above the surface of gravel, given that the flow rate (q) is to be maintained at 40 liters per second.

Quiz # 1 q1 = q2 = q = 4010-3 m3/s a1 = a2 = a = r2 = (0.75)2 = 1.767 m2 v1 = v2 = v = q/a = 2.26410-2 m/s i1 = v1/k1 = 2.26410-2/1010-2 = 0.2264 i2 = v2/k2 = 2.26410-2/110-2 = 2.264 i1 = 0.2264 = h1/L1 = (H + 1 – z)/(1.0) i2 = 2.264 = h2/L2 = (z + 1)/(1.0) Solving the equations for z and H  z = 1.264 m H = 0.4904 m

Quiz # 1

Quiz # 2 For a clay deposit: Cc = 0.4 and Cr = 0.05, determine the final void ratio for each of the following loading conditions: a) Initially s¢ = 50 kPa, e = 1.2, Finally s¢ = 90 kPa, and given that s¢pc = 100 kPa b) Initially s¢ = 50 kPa, e = 1.2, Finally s¢ = 190 kPa, and given that s¢pc = 100 kPa

Quiz # 2 C = slope = e/logs’ = (ei – ef)/log(s’f/s’i) (a) 0.05 = (1.2 – ef)/log(90/50)  ef = 1.187 (b) Two stages: 0.05 = (1.2 – efi)/log(100/50)  efi = 1.185 0.4 = (1.185 – ef)/log(190/100)  ef = 1.073