Download pdf - Chapter 5-Shear Strength

8/11/2019 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