Chapter 5-Shear Strength

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    5.1 SHEAR STRENGTH

    CLO 1 :Discover the physical & mechanical

    properties of soils basically for engineering practice

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    LEARNING LESSON OUTCOME:

    1. Understand shear strength of soil.

    2. Understand Mohr-Coulumb criteria for failure

    of soil.

    3. Understand the parameter of shear strength.

    4. Know the tri-axial methods

    At the end of this chapter, students will be able

    to:

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    DEFINITION:

    Shear strength of soil is the maximum value of shear stress

    that may be induced within its mass before the soil yields.

    WHAT WE ARE GOING TO FIND IN THIS CHAPTER?

    The value of:

    = angle of friction (geseran)

    C = cohesion (kejelekitan)

    TYPES OF SHEAR TEST

    i)Shear box test

    ii)Triaxial compression test

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    Types of graph The condition of soil

    -For the non-cohesion soil, soil friction or

    dry soil, or saturated soil.

    -c= 0 but have a value of

    -For the cohesion soil or saturated clay soil.

    - = 0 but have a value of c

    -For the sandy clay silty clay

    - Have a value of = 0 and c

    c

    c

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    Example 1:

    A drained shear box test was carried out on a sandy clay and

    yielded. The result as below:

    Vertical load (kg) Dial gauge reading at proving

    ring (1 part = 1m)

    36.8 17

    73.5 26

    110.2 35

    146.9 44

    The size of shear box is 60mm x 60mm and coefficients is 20N/

    1m.

    Determine the shear strength parameters for this soil.

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    Solution:

    Vertical load

    (M)

    = N/A

    (kN/m2)

    Dial gauge

    reading (S)

    = F/A

    (kN/m2)

    36.8 17

    73.5 26

    110.2 35

    146.0 44

    = F/A = [20 (S) / A]

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    Vertical load (kN) Shear load at failure (kN)

    0.36 0.32

    0.72 0.52

    1.08 0.70

    1.44 0.88

    Example 2:

    The result below was collected from shear box test on the

    sample with a dimension 60mm x 60mm. Determine the value ofshear strength parameters.

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    Solution:

    Vertical load

    (kN)

    Shear load at

    failure (kN)

    = N/A

    (kN/m2)

    = F/A

    (kN/m2)

    0.36 0.32 100 88.9

    0.72 0.52 200 144.4

    1.08 0.70 300 194.4

    1.44 0.88 400 244.4

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    Example 3:

    A drained triaxial compression test carried out on three

    specimens of the same soil yielded the following results:

    Test No. 1 2 3

    Cell pressure (kN/m2) 100 200 300

    Deviator stress at failure (kN/m2) 210 438 644

    Find value of shear strength parameters.

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    Solution:

    The principal stresses:-

    Minor principal stress, 3= r = cell pressure

    Major principal stress, 1= a= cell pressure + deviator stress

    since uf= 0, Then , 1= 1and 3= 3

    Test No. 1 2 3

    '3(kN/m2) 100 200 300

    1 (kN/m2) 310 638 944

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    Example 4:

    The result below was collected from triaxial test. Determine

    the parameter value.

    CELL PRESSURE

    (kN/m

    2

    )

    DEVIATOR STRESS

    (kN/m

    2

    )

    PORE WATER

    PRESSURE (kN/m

    2

    )150 192 80

    300 341 154

    450 504 222

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    Solution:

    Minor principal stress, 3= 3 = cell pressure

    Major principal stress, 1= 1= cell pressure + deviator stress

    Do you still remember this , = - u

    3 1

    70 262

    154 487

    228 732