     • date post

23-Oct-2020
• Category

Documents

• view

25

1

Embed Size (px)

Transcript of SOIL MECHANICS & FOUNDATION ENGINEERING SOIL MECHANICS & FOUNDATION ENGINEERING ECA Page 95...

• SOIL MECHANICS & FOUNDATION ENGINEERING ECA

Page 95

SOIL MECHANICS & FOUNDATION ENGINEERING

2016

1. A sample of normally consolidated clay was subjected to a consolidated undrained triaxial compression test that was carried out until the specimen failed at a deviator stress of 50 kN/m2. The pore water pressure at failure was recorded to be 20 kN/m2 and confining pressure of 50 kN/m2 was used in the test. Determine the consolidated undrained friction angle. (15 marks) Deviator stress = 50 kN/m2, Pore water pressure(σu ) = 20 kN m2⁄ , Confining Pressure (σ3)failure = 50 kN/m2 Since the given test is consolidated undrained triaxial compression test (σ�3)failure = (σ3)failure − σu = 50 − 20 = 30 kN/m2 ∴ (σ1)failure = (σ�3)failure + deviator stress (σ1)failure = 30 + 50 = 80 kN/m2 Using the formula:

Sin ϕ = (σ1)failure − (σ�3)failure(σ1)failure + (σ�3)failure

Sin ϕ = 80−30 80+30

ϕ = sin−1 � 50 110

ϕ = 2702′ 2. State and discuss different factors influencing compaction of soil in the field. (20 marks) Factors influencing compaction of soil in the field are as follows: (a) Moisture content: At low water content, soil is stiff and offers more resistance to compaction. As moisture content increases, a film of water surrounds the soil particles which tends to lubricate the particles and make them easier to be worked around, hence they come more closer and become dense. This phenomena occurs upto OMC beyond which water starts occupying the space previously occupied by soil solids. As γw < γsolids ; dry unit weight decreases. (b) Compactive Effort: For a given type of compaction, higher the compactive effort, higher is the maximum dry unit weight and lower is the OMC. As shown in the figure, the compaction curve shifts to the top and to he left when compactive effort is increased.

• SOIL MECHANICS & FOUNDATION ENGINEERING ECA

Page 96

(c) Type of Soil:  Coarse grained soil if well graded is compacted to high 𝛾𝛾𝑑𝑑 especially if they contain some fines.

However, if the quantity of fines is excessive, 𝛾𝛾𝑑𝑑 decreases.  Poorly graded or uniformly graded sand lead to lowest dry unit weight values.  In clay soil, maximum dry unit weight tends to decrease as plasticity increases.  Heavy clay with high plasticity has very low dry unit weight and very high OMC. (d) Method of compaction used in the field: Several methods are used for compaction of soil in field. The choice of method will depend upon the soil type, the maximum dry density required and economic consideration. Some of the commonly used compaction methods are as follows:  Tampers  Rollers

Rollers may be of following types: • Smooth wheel rollers • Sheep foot roller • Pneumatic tyred rollers • Vibratory compactors

 Some other methods of compactions are also listed below: • Vibro Floatation method • Tera Probe method • Compaction by Pounding • Compaction Piles • Compaction by explosives

3. A retaining wall with a smooth vertical back is 9 m high and retains a two layer sand backfill with the following properties 0 – 3 m depth: c’ = 0.0, ф = 30°, 𝛄𝛄 = 18 kN/m3

3 – 9 m depth: c’ = 0.0, ф = 35°, 𝛄𝛄 = 20 kN/m3

• SOIL MECHANICS & FOUNDATION ENGINEERING ECA

Page 97

Show the active earth pressure distribution and determine the total active thrust on the wall. Assume that the water table is well below the base of the wall. (20 marks)

ACTIVE EARTH PRESSURE

pa1 = ka1γ1H1 = � 1 3

× 18 × 3� = 18 kN/m2

pa2 = ka2γ1H1 = (0.271 × 18 × 3) = 14.634 kN/m 2

pa3 = ka2γ2H2 = (0.271 × 20 × 6) = 32.52 kN/m 2

ACTIVE THRUST ON THE WALL

Pa1 = ka1γ1H1

2

2 =

1 3 × 18 × 3

2

2 = 27 kN/m

Pa2 = ka2γ1H1H2

2 = 0.271 × 18 × 3 × 6

2 = 87.804 kN/m

Pa3 = ka2γ2H2

2

2 =

1 3 × 18 × 3

2

2 = 97.56 kN/m

TOTAL ACTIVE THRUST ON THE WALL Pa = Pa1 + Pa2 + Pa3 = 27 + 87.904 + 97.56 = 212.364 kN/m 4. A layer of sand 6.0 m thick lies above a layer of clay soil. The water table is at a depth of 2.0 m below the ground surface. The void ratio of the sand layer is 0.6 and the degree of saturation of the sand layer above the water table is 40%. The void ratio of the clay layer is 0.7. Determine the total stress, neutral stress and effective stress at a point 10 m below the ground surface. Assume specific gravity of the sand and clay soil respectively as 2.65 and 2.7. (20 marks) Given: For sand layer e = 0.6, for clay layer e = 0.7, G (sand) = 2.65, G (clay) = 2.70. Let γw = 9.81 kN/m3 Let us divide the layers in 3 sections: For 1 section:

γbulk = (G+Se )γw

1+e

• SOIL MECHANICS & FOUNDATION ENGINEERING ECA

Page 98

γbulk = (2.65+ 0.4 × 0.6)9.81

(1+0.6) = 17.72 kN/m3

For 2 Section:

γsat = (2.65+ 0.6)9.81

(1+0.6) = 19.93 kN/m3

For 3 section i.e for clay:

γsat = (2.70+ 0.7)9.81

(1+0.7) = 19.62 kN/m3

Total pressure (𝝈𝝈), Pore Water pressure (𝝈𝝈𝐮𝐮) and Effective stress (𝝈𝝈�) At A: σ = γbulk × 0 = 0 σu = 0 σ� = σ − σu = 0

At B: σ = γbulk × 2 = 17.72 × 2 = 35.44 kN/m2 σu = 0 σ� = σ − σu = 35.44 kN/m2

At C: σ = (γbulk × 2) + (γsat . × 6) = (17.72 × 2) + (19.93 × 6) = 155.02 kN/m2 σu = 9.81 × 6 = 58.86 kN/m2 σ� = σ − σu = 155.02− 58.86 = 96.16 kN/m2 At D: σ = (γbulk × 2) + (γsat . × 6) +(γsat )clay × 2 = (17.72 × 2) + (19.93 × 6) + (19.62 × 2) = 194.26 kN/m2 σu = 9.81 × 8 = 78.48 kN/m2 σ� = σ − σu = 194.26− 78.48 = 115.78 kN/m2

• SOIL MECHANICS & FOUNDATION ENGINEERING ECA

Page 99

Points Total Pressure (𝛔𝛔) Pore water Pressure (𝛔𝛔𝐮𝐮)

Effective stress(𝛔𝛔�) 𝛔𝛔� = 𝛔𝛔 − 𝛔𝛔𝐮𝐮

A 𝜎𝜎 = 0 0 0 B 𝜎𝜎 = 35.44 0 35.44 C 𝜎𝜎 = 155.02 58.86 96.16 D 𝜎𝜎 = 194.26 78.48 115.78

2015

5. The soil from a borrow area having an average in-situ unit weight of 15.5 kN/m3 and water content of 10%, was used for the construction of an embankment (total finished volume 6000 m3). In half of the embankment due to improper rolling, the dry unit weight achieved was slightly lower. If the dry unit weights are in the two parts are 16.5 kN/m3 and 16.0 kN/m3, find the volume of borrow area soil used in each part and the amount of soil used. (15 marks)

• SOIL MECHANICS & FOUNDATION ENGINEERING ECA

Page 100

Borrow Pit Embankment Total volume = 6000 m3

γbulk = 15.5 kN/m2 w = 10 % Total volume = 𝑥𝑥 m3 Let weight of the soil in borrow pit be W1

Half Embankment Volume (V1) = 3000 m3 γbulk = 16 kN/m3 Weight of the soil = 16 × 3000 = 48000 kN

Half Embankment Volume (V2) = 3000m3 γbulk = 16.5 kN/m3 Weight of the soil = 16. 5 × 3000 = 49500 kN

Combined weight of the soil(W2) = 97500 kN NOTE: Weight of the soil does not changes so W1 = W2 For the borrow pit, W2 = γbulk x volume of borrow pit

Volume of Borrow Pit = W 2 γbulk

= 97500 15.5

Volume of Borrow Pit = 6290.32 m3 6. A 6.0 m high retaining wall is to support a soil with unit weight 17.4 kN/m3, ϕ = 26° and c’ = 14.36 kN/m2. Determine the Rankine active force per unit length of the wall before the tensile crack occurs. Find the critical depth. (15 marks)

Given: Height of retaining wall = 6 m, γbulk = 17.4 kN/m3, ϕ = 26° , c’ = 14.36 kN/m2

ka = cot2 �45 + ϕ 2 �

ka = cot2 �45 + 26 2 �

∴ ka = 0.39 Rankine’s active earth pressure on unit length, Pa =

kaγH2

2 − 2cH�ka

Pa = 0.39 × 17.4 × 62

2 − 2 × 14.36 × 6 √0.39

Pa = 122.148− 107.61

• SOIL MECHANICS & FOUNDATION ENGINEERING ECA

Page 101

Pa = 14.538 kN/m Total net vertical pressure at the depth of 2𝑧𝑧0 in cohesive soil is 0, hence these cohesive soils can stand with their vertical face up. This depth is known as critical depth of unsupported cut

z0 = 2c ′

γ�ka = 2 × 14.36

17.4 √0.39

z0 = 2.643 m

2z0 = 2 × 2.643 = 5.286 m

2014

7. A soil sample in its natural state has, when fully saturated, a water content of 32.5%. Determine the void ratio, dry and total unit weights. Calculate the total weight of water required to saturate a soil mass of volume 10 m3. Assume GS = 2.69. (15 marks)

Void ratio: From S = 𝑤𝑤Gs 𝑒𝑒

∴ 𝑒𝑒 = 𝑤𝑤Gs S

= 32.5 × 2.69 (1) × 100

= 0.874

Total unit length: F