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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 = 80−3080+30

ϕ = sin−1 � 50110

�

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.


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(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


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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 = �13

× 18 × 3� = 18 kN/m2

pa2 = ka2γ1H1 = (0.271 × 18 × 3) = 14.634 kN/m2

pa3 = ka2γ2H2 = (0.271 × 20 × 6) = 32.52 kN/m2

ACTIVE THRUST ON THE WALL

Pa1 =ka1γ1H1

2

2=

13 × 18 × 3

2

2 = 27 kN/m

Pa2 =ka2γ1H1H2

2= 0.271 × 18 × 3 × 6

2 = 87.804 kN/m

Pa3 =ka2γ2H2

2

2=

13 × 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


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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


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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)


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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 2γbulk

= 9750015.5

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 +ϕ2�

ka = cot2 �45 +262�

∴ 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


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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 = 𝑤𝑤Gs𝑒𝑒

∴ 𝑒𝑒 = 𝑤𝑤GsS

= 32.5 × 2.69(1) × 100

= 0.874

Total unit length: From

𝛾𝛾𝑡𝑡 = Gsγw (1+w)

1+𝑒𝑒

= 2.69(9.81)(1 + 0.325)

1+0.874

= 18.7 kN/m3

Dry unit weight: From

𝛾𝛾𝑑𝑑 = γw Gs1+𝑒𝑒

= 2.69 × 9.81

1+0.874

= 14.08 kN/m3

From, W = 𝛾𝛾𝑡𝑡V = 18.66 × 10 = 186.6 kN From, WS = 𝛾𝛾𝑑𝑑V = 14.08 × 10 = 140.8 kN Weight of water = W – WS = 186.6 – 140.8 = 45.8 kN

2013

8. Find out the time required for 50 % consolidation in a soil having thickness of 800 cm and pervious strata at top and bottom. What will be the value of coefficient of consolidation if coefficient of permeability = 0.0000001 cm/sec? Void ratio = 1.8, 𝐦𝐦𝐯𝐯 R = 0.0008 𝐜𝐜𝐦𝐦𝟐𝟐/𝐠𝐠𝐦𝐦, Time factor (𝐓𝐓𝐯𝐯) = 0.3, 𝜸𝜸𝒘𝒘 = 1 gm/cc (15 marks)


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Time required for 50% consolidation k = 0.0000001 cm/sec Void ratio, e = 1.8 mv = 0.0008 cm2/gm Time factor (Tv) = 0.3 ρwater R = 1 gm/cm3

Coefficient of consolidation, Cv Cv =

kmvγw

Cv = 0.0000001(.0008) × 1

Cv = 1.25 × 10−4 cm2/sec

Tv = CV . t

d2

∴ t = TV .d2

CV �d = 800

2 = 400 cm�

t = 0.3 × (400)2

1.25 × 10−4 (two way drainage)

t = 384 × 106 sec

t = 12.1766 years 9. Calculate the ultimate bearing capacity per unit area of: (15 marks) (i) A strip footing 1 m wide; (ii) A square footing 3 m × 3 m; (iii) A circular footing of diameter 3 m

Given: Unit weight of the soil is 1.8 t/m3, cohesion = 2 t/m2 and ϕ = 200. Nc = 17.5, Nq = 7.5 and Nr = 5 (i) A strip footing 1 m wide A strip footing is 1 m wide.

qu = CNc + γDfNq + 12

γBNr

qu = (2 × 17.5) + (1.8 × 0 × 0.75) + �12

× 1.8 × 1 × 5�

qu = 39.5 t/m2 (ii) A square footing 3 m × 3 m

qu = 1.3 CNc + γDfNq + 0.4 γBNr

qu = (1.3 × 2 × 17.5) + (0.4 × 1.8 × 3 × 5) qu = 56.3 t/m2

(iii) A circular footing of diameter 3 m

qu = 1.3 CNc + γDfNq + 0.3 γDNr

qu = 1.3 CNc + 0.3 γDNr = (1.3 × 2 × 17.5) + (0.3 × 1.8 × 3 × 5)


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qu = 53.6 t/m2

2012

10. A sand deposit is 10 m thick and overlies a bed of soft clay. The ground water table is 3 m below the surface. If the sand above the ground water table has a degree of saturation of 45%, plot the diagram showing the variation of the total stress, pore water pressure and the effective stress. The void of the sand is 0.70. Take G = 2.65. (20 marks)

γt sand = G + Se1 + e

γw

= 2.65 + 0.45 × 0.71+0.7

× 10 = 17.44 kN/m3

Unit wt. of saturated sand γsat (sand )

= G+Se1+e

γw

= 2.65+ 0.71+0.7

x 10 = 19.71 kN/m3

At elevation – 3 m 𝜎𝜎 = 𝑧𝑧γtt = 3 × 17.44 = 19.71 kN/m2 At elevation – 10 m 𝜎𝜎 = 3γt + 7 γsat = 3 × 17.44 + 7 × 19.71 = 190.29 kN/m2 V = 7 × γw = 70 kN/m2 𝜎𝜎 = 190.29− 70 = 120.29 kN/m2


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2011

11. In a consolidation test on a soil, the void ratio of the sample decreases from 1.24 to 1.12 when the pressure is increased from 20 to 40 tonnes/sq. m. Calculate the co-efficient of permeability of the soil during this pressure increment is 8.5 × 𝟏𝟏𝟏𝟏−𝟑𝟑 cm/sec. (15 marks) Given ρ0 = 1.24, ρ1 = 1.12, σ1 = 20 t/m2, σ2 = 40 t/m2 Coefficient of permeability = 8.5 × 10−3 cm/sec Cv =

kγw mv

, mv = av

1+ ρ0

When mv = coefficient of volume change av = coefficient of compressibility Cv = ? av =

1.24−1.1240−20

= 6 × 10−3 m2/tonne

mv = av

1+ ρ0= 6 × 10

−3

1+1.24 = 2.68 × 10−3 m2/tonne

Cv = 8.5 × 10−3 cm /sec

9.81 kN /m2 × 2.68 × 10−3 m2/tonne

= 8.5 x 10−3

100 m/year × 365 × 24 × 60 × 60

9.81 kN /m2 × 2.68 × 10−3 m2

�1000 ×9.811000 �kN

= 8.5 × 10−3 × 365 × 24 × 60 × 60 × 9.81

100 × 9.81 × 2.68 × 10−3

Cv = 1000208.95 m2/year

12. In a plate bearing test on pure clayey soil failure occurred at a load of 12.2 tonnes. The size of the plate was 45 cm × 45 cm and the test was done at the depth of 1.0 m below ground level. Find out the ultimate bearing capacity for a 1.5 m wide continuous wall footing with its base at a depth of 2 m below the ground level. The unit weight of clay may be taken as 1.9 gm/cc and Nc = 5.7 and Nq = 1 and Nr = 0. (20 marks)


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Plate bearing capacity of plate

Qup = P

Ap= 12.2 × 9.81

0.452

Qup = 591.02 kN/m2 In clay Qup = Quf Hence, the ultimate bearing capacity of foundation Quf = 591.02 kN/m2 Qu = cNc + γDfNq +

12γBNr

Nr = 0 591.02 = C × 5.7 + 1.9 × 9.81 × 2 × 1

⇒ C = 97.148 kN/m2 13. Write a short note on significant properties of soil. (10 marks) Significant properties of soil are: (a) Texture: It refers to the relative proportions of particles of various sizes such as sand, silt and clay in

the soil. (b) Structure: It refers to the aggregation of primary soil particles (sand, silt and clay) into compound

particles or cluster of primary particles which are separated by the adjoining aggregates by surfaces of weakness. Structure modifies the effect of texture in regard to moisture and air relationships, availability of nutrients, action of microorganisms and root growth.

(c) Particle density: Particle density of soil is the mass per unit volume of soil particles (soil solid phase) expressed in gm/cc. Most soils have particle density of about 2.6 g/cc. Presence of organic matter decreases the density and iron compounds increase the density.

(d) Bulk density: Bulk density of soil is the mass of soil per unit volume of soil (Volume of soil includes both soil solids and pores).

(e) Soil Colour: Soil colour gives an indication of the various processes going on in the soil as well as the type of minerals in the soil. For example, the red colour in the soil is due to the abundance of iron oxide under oxidized conditions whereas dark colour is generally due to the accumulation of highly decayed organic matter.

(f) Soil Permeability: It is the ability of the soil to transmit water and air. An impermeable soil is good for aquaculture as the water loss through seepage or infiltration is low. As the soil layers or horizon s vary in their characteristics, the permeability also differs from one layer to another.

(g) Particle size distribution: It is a method of separation of soil into different fraction based on the particle size.

(h) Grain Shape: Soil grains are called bulky incase of sand and gravel which are further classified as : angular, subangular, subrounded, rounded, well rounded. Higher angularity of soil grain leads to higher shearing strength.

(i) Relative density or Density Index (ID) : It represents degree of denseness or coarseness of natural deposits of coarse grained soil. Higher relative density means higher shear strength and low compressibility and vice –versa. ID =

emax − e naturalemax − emin

× 100

(j) Atterberg’s Limit and Consistency: Consistency represents the relative ease with which a soil can be deformed.


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Consistency is related to water content i.e., how with change in water content, the consistency of the soil changes. Atterberg classified the consistency in four stages. The behavior of the soils different indifferent stages:

• Solid stage • Semi solid stage • Plastic stage • Liquid stage

2010

14. The following properties of the soil were determined by performing tests on clay sample: − Natural moisture content = 25% − Liquid limit = 32% − Plastic limit = 14% − Diameter of 60% size = 0.006 mm − Diameter of 10% size = 0.006 mm Calculate the (i) liquidity coefficient, (ii) uniformity coefficient and (iii) relative consistency. (15 marks) Given, Natural moisture content w = 25% LL or WL = 32, WP = 34, D60 = 0.006 mm, D10 = 0.006 mm (i) Liquid consistency

IL = w− Wp

W L − WP × 100

IL = 0.25−0.240.32−0.24

× 100

IL = 12.5%

(ii) Uniformity coefficient

Cu = D60D10

= 0.0060.006

Cu = 1 (iii) Relative consistency

IC = W L− w

W L − W P × 100

IL = 0.32−0.250.32−0.24

× 100

IL = 87.5%

15. A sample of soil 10 cm diameter, 15 cm length was tested in a variable head permeameter. The initial head of water in the burette was found was found to be 45 cm and it was observed to drop to 30 cm in 195 seconds. The diameter of the burette was 1.9 cm. Calculate the coefficient of permeability in meter/day. (15 marks)


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Given, D = 10 cm, L = 15 cm, A =

π4

× D2 = π4

× 102 = 78.53 cm2 Initial head of water in burette h1 = 45 cm Drop in head = 30 cm Final head h2 = 45 – 30 = 15 cm Time observed to drop = 195 sec Dia. of burette = 1.9 cm Area of burette A =

π4

× 1.92 = 2.835 cm2 So the coefficient of permeability by variable head permeability test is

K = 2.303 log10�

4515� × 2.835 × 15

78.53 × 195

= 3.05 x 10−3 cm/sec

= 3.05 x 10−3

100× 24 × 60 × 60 m/day

= 2.636 m/day

16. Explain standard penetration test for measuring the penetration resistance of soil. (12 marks) Standard Penetration Test (SPT) is suitable for dense sand. In loose saturated liquefaction may occur

& in clays remoulding may occur or pore pressure resistance may develop. Test is performed either at 2 – 2.5 m depth interval or at change of stratum A bore hole is made at the test level and split spoon sampler is placed on the bore hole. The test is performed in 3 stages, 150 mm penetration in each stage The SPT no. is taken as number of blows required for last 300 mm penetration; i.e. numbers of

blows for first 150 mm penetration are ignored. Standard weight of hammer is 65 kg and standard height of fall is 75 cm.

Important: The observed value of SPT no. is subjected to following two corrections applied in sequence: (a) Overburden pressure correction: If No is observed reading of SPT, then corrected value of SPT for overburden pressure is given by

N1 = N0 ×350σ�+70

𝜎𝜎 � is effective overburden pressure at test level in kN/m2 σ� ≤ 280 kN/m2

If σ� ≥ 280 kN/m2, then overburden pressure correction is not required, i.e. N1 = No

(b) Water Table correction or Dilatancy correction: The corrected value for overburden pressure is further corrected for presence of water table. In case of fine sands and silts the pore pressure may have apparent resistance therefore correction is required. This correction is required only when water table is at or above test level If N2 is corrected value for dilatenccy, then N2 = 15 +

12

(N1 − 15) And, if N1 < 15; no correction is required.


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17. An earthen embankment is compacted to a dry density of 1.82 gm/cc at moisture content of 12%. The bulk density and moisture content are 1.72 gm/cc and 6% at the site is borrowed and transported at the site of construction. How much excavation should be carried out in the pit of borrowed area for each cu-m of the embankment? (18 marks)

At site from where soil is borrowed Vt = 1.72 gm/cc or 1720 kg/m3, m/c = 6% Vd = 1.82 gm/cc, m/c = 12% Volume of embankment soil = 1 m3

Dry density Vd = W sV

Given, Vd = 1.82 gm/cc = 1.82

1000× 106 kg/m3

= 1820 kg/m2 Wt. of solids from each cu-m of embankment Ws = 1820 × 1 = 1820 kg As we know, while coring soil from borrow pit to earthen embankment wt. of solids remain same, hence, wt. of solids at site = 1820 kg

Unit wt. of soil at site Vt = W V

Also W = W WW S

W/C or m/c = 6% given

0.06 = W W1820

⇒ WW = 1820 ×

6100

= 109.2 kg Total wt. of soil at site W = WW + Ws = 109.2 + 1820 = 1929.2 kg

Vt = 1720 = 1929.2

V

Volume of soil to be excavated from borrow pit V = 1929.21720

= 1.212 m3


Page 109

2009

18. A sample of soil has a porosity of 35 percent and specific gravity of solids is 2.67. Calculate void ratio, dry density and unit weight, if (i) the soil is 50% saturated (ii) the soil is 100% saturated = 1 (15 marks) (i) the soil is 50% saturated Given, n = 0.35, G = 2.67, e = ?, ρd= ?, γ = ?, w = 0.5

γ = �G + Se1 + e

� 𝛾𝛾𝑤𝑤 = �2.67+ 0.5 × 2.67

1+ 0.2593�× 1

γ = 3.18 void ratio e =

n1−n

= 0.351−0.35

= 0.54

dry density ρd = �Gρ

1+e� = 2.67× x 9.8

1+0.54

= 17 KN/m3 If the soil is 50% saturated Sr = 50% = 0.50

e = 𝑤𝑤GSr

⇒ 𝑤𝑤 = 𝑒𝑒sr

G

= 0.54 × 0.5

2.67 = 0.101

γ = γd (1 + w) = 17 × (1 + 0.101) = 18.72 (ii) the soil is 100% saturated = 1

γ = �G+Se1+e

� 𝛾𝛾𝑤𝑤 = �(2.67+ 0.54 × 1.0) × 9.81

1+0.54�

= 20.45 kN/m3 19. A sample of soil is 5 cm high and 8 cm in diameter. It was tested in a constant head permeameter. Water perlocates through the soil under a constant head of 45 cm for 8 m. The water was collected and weighed. Its weight was recorded as 500 gm. On oven drying the sample of soil, the weight was recorded as 450 gm. If G is 2.65, calculate (i) coefficient of permeability (ii) seepage velocity of water when the water was under operation. (15 marks) (i) coefficient of permeability Given, Height of soil sample L = 5 cm, dia D = 8 cm then, A =

π4

× 82 = 50.26 cm2 head of water, h = 45 cm, volume of soil sample = A × L = 50.26 × L = 50.26 × 5 = 251.33 cm3 Wt. of water recorded = 500 gm


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Vol. of water = wt. of water/unit wt. of water

= 5009.81

= 50.968 m3 Permeability of soil by constant head permeability method

K = QL

A𝑡𝑡ℎ= qL

Aℎ= 50.968 × 5

50.26 × 45

= 0.11 cm/sec … �Q𝑡𝑡

= 50.968 m3� (ii) seepage velocity of water when the water was under operation. Seepage velocity (vs) = kp i i = head of water/length of soil sample

= 455

= 9

kp = kn

Dry unit weight = dry wt. of soil sample/vol. of soil sample

= 450

251.33cm 3= 1.79 gm/cc

yd = Gγw1+e

= 2.65 × 11+e

e = 0.48

n = e

1+e= 0.48

1.48= 0.32

kp = kn

= 0.481.48

= 0.34

vs = kp × i = 0.34 × 9 = 3.05 m/sec 20. An embankment was compacted at moisture content of 15%. Its density was determined with the help of a core cutter and the following data was collected. − Empty weight of the cutter = 1200 m − Weight of cutter when it is full of soil = 3200 gm − Volume of the cutter = 1000 cc Calculate bulk density and saturation percentage of the embankment becomes fully saturated due to rains, the, determine its moisture content and saturated density. Take G = 2.70. (20 marks) Given, Mass of soil in cutter = 3200 – 1200 = 2000 g Bulk density ρ = M/V = 2000/1000 = 2 g/cm3 Bulk unit weight γ = 9.81 × ρ = 9.81 × 2 = 19.62 kN/m3

γd = γ

1+𝑤𝑤= 19.62

1+0.15

= 17.06 kN/m3

e = Gγwγd

− 1

= 2.7 × 9.81

17.06− 1


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= 0.553

Sr = 𝑤𝑤 .G𝑒𝑒

= 0.15 × 2.70.553

= 0.732 = 73.2% At saturation: Since the volume remains the same, the voids ratio also remains unchanged. e = 𝑤𝑤sat .G 𝑤𝑤sat . = e/G = 0.553/2.7 = 0.2048 = 20.48%

γsat . = (G+e)γw

1+e

= (2.7+0.553)9.81

1+0.533

= 20.55 kN/m3 21. Explain the factors which affect the bearing capacity of soils. (10 marks) The bearing capacity of foundation is the maximum load per unit area which the soil can support without failure. It depends upon the shear strength of soil as well as shape, size, depth and type of foundation. All civil engineering structures whether they are buildings, dams, bridges etc. are built on soils. A foundation is required to transmit the load of the structure on a large area of soil. The foundation of the structure should be so designed that the soil below does not fail in shear nor there is excessive settlement of the structure. The conventional method of foundation design is based on the concept of bearing capacity. Factors affecting Bearing capacity of Soils: (a) Type of soil: The bearing capacity of soils depends upon the type of soil. Terzaghi bearing capacity equation for strip footing:

qu = CNc + qNq + 0.5γBNγ For cohesionless soil like sands qu = qNq + 0.5γBNγ For clay, qu = CNc + qNq (b) Physical characteristics of foundation: Physical characteristics like width, shape and depth of foundation affects the bearing capacity of soils. The bearing capacity of soils depends upon the width B and depth D of foundation. The shape of the foundation also affects the bearing capacity. (a) For square footings:

qu = 1.3CNc + qNq + 0.4γBNγ (b) For circular footings:

qu = 1.3CNc + qNq + 0.3γDNγ (c) Soil Properties: Soil preoperties like shear strength, density, permeability etc., affect the bearing capacity of soil. Dense sand will have more bearing capacity than loose sand as unit weight of dense sand is more than loose sand.


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(d) Type of foundations: The type of foundation selected also affects the bearing capacity of soils. Raft or mat foundation adopted supports the load of structure safely by spreading the load to a wider area, even if the soil is having low bearing capacity (e) Water table: When the water is above the base of the footing, the submerged unit weight of the soil is used to calculate the overburden pressure and the bearing capacity of the soil is reduced. (f) Amount of Settlement: The amount of settlement of the structure also affects the bearing capacity of soil. If the settlement exceeds the possible settlement, the bearing capacity of soil is reduced. (g) Eccentricity of Loading: If the load acts eccentrically in a footing the width ‘B’ and length ‘L’ should be reduced as under

B’ = B – 2e L’ = L – 2e and A’= B’ x L’

The ultimate bearing capacity (qu ) of such footings are determined by using B’ and L’ instead of B and L. Hence qu is less than that corresponds to actual size of footing.

2008

22. A sample of dry soil having specific gravity of 2.74 and having a mass of 133.7 gm is uniformly dispersed in water to form 1000 cc of suspension. (i) Determine the density of suspension immediately after it is prepared. (ii) A 10 cc of the suspension was removed from the depth of 21 cm beneath the top surface after the suspension was allowed to stand for 2 min 30 sec. The dry mass of the soil in the sample drawn was found to be 0.046 gm. Determine one point on the grain-size distribution curve corresponding to this observation. − Temperature of suspension = 20 ℃ − Viscosity of water at 20 ℃ = 0.0102 poise (15 marks)

𝑒𝑒𝑤𝑤 = 1 g/cc Vt =

951.2 + 133.71000

Vt = 1.0849 g/cc


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Setting velocity according to Stroke’s law Vs =

( Vs − Vw )D2

18μ= g(ρs − ρw )D

2

18μ

= gρw (Gs − 1)D2

18μ

Vs = g (Gs − 1)D2

18ϑ � μ

ρw= ϑ�

Also, Vs = He

t= 10

120+30= 10

150 cm/sec

(As std. ht. of sample collection is 10 cm in Apette)

10

150= 9.81 × 100 (2.74 − 1)D

2

18 × 0.0102

D = 0.00268 cm = 0.0268 mm

% finer = (0.406/10)

(133.7/1000) = 0.3036 or 30.36%

23. 60 cm diameter well is being pumped at a rate of 1360 liters/minute. Measurements in a nearby test well were made at the same time as follows. At a distance of 60 m from the well being pumped, the drawdown was 6 m and at 15 m, the drawdown was 1.5 m. The bottom of the well is 90 m below the ground water table. (i) Find out the coefficient of permeability. (ii) If all the observed points were on the Dupuit curve, what was the drawdown in the well during pumping? (iii) What is the specific capacity of the well? (iv) What is the rate at which water can be drawn from this well? (15 marks) (i) Find out the coefficient of permeability. According to Thiem’s formula for unconfined aquifers,

Q = πK�h22− h12�

2.3log 10r2r1

Here, r1= 6 m; r2 = 15 m s1 = 6 m; s2 = 1.5 m d = 90 cm Q = 1360 litres/minute = 1.36 m3/min h1 = 90 – 6 = 84 m h2 = 90 – 1.5 = 88.5

1.36 = πK�88.5 2− 84 2�

2.3log 10156

or πK = 1.36 × 2.3 × 0.398172.5 × 4.5

or πK = 1.603 × 10−3 and K = 0.51 × 10−3 m/min (ii) If all the observed points were on the Dupuit curve, what was the drawdown in the well during

pumping? Now, rw = 0.3 m; r2 = 15 m


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or h2 = 88.5 m; hw = ?

Using Q = πK�h22− hw2 �

2.3log 10r2rw

We get, Q = πK�88.5 2− hw2 �

2.3log 10150.3

But Q = 1.36 m3/min and πK = 1.603 × 10−3 ∴ 1.36 = 1.603 x �10

−3� × �88.52− hw2 �2.3log 10 50

or 1.36 ×2.3 × 1.69

1.603 × 10−3 = 88.52 − hw2

or 3290 = 7820 – hw2 or hw2 = 7820 – 3290 = 4530

or hw2 = 67.4 m ∴ Drawdown in the pumped well = 90 – 67.4 = 22.6 m (iii) What is the specific capacity of the well? Specific capacity of the well is the discharge for a unit (i.e. 1 m) drawdown in the pumped well. Let us first find out the value of R Use Dupuit’s equation for unconfined aquifers, i.e. eq. as

Q = πK�d2− hw

2�2.3log 10 R/0.3

∴ 1.36 = πK�902− 67.42�

2.3log 10 R/0.3

or 1.36 = 1.603 x �10−3� × 157.4 x 22.6

2.3log 10 R/0.3

log10 R/0.3 = 1.603 × 157.4 × 22.6

2.3 × 1360

Taking antilog, we get R

0.3 = 66.7

R = 20.01; say R = 20 m Now, specific capacity = Qunit drawdown

= πK�902− 892�

2.3log 10 20/0.3

= 1.603 × �10−3� × 179 × 1

2.3 × 1.824

= 68.3 × 10−3 m3/min

Hence, the specific capacity = 68.3 litres/minute (iv) What is the rate at which water can be drawn from this well? Maximum discharge will occur when hw = 0


Page 115

∴ Qmax = πK�902− 02�

2.3 log 10 20/0.3

= 1.603 × 10−3 x 8100

2.3 × 1.824

= 3.09 m/min Hence, the maximum rate of discharge = 3090 l/m. 24. A direct shear box test performed on a remoulded sand sample yielded the following observations at the time of failure: − Normal load = 0.36 kN − Shear load = 0.18 kN − The sample area was 36 cm2. Determine: (i) the angle of internal friction, (ii) the magnitude and direction of the principal stresses in the zone of failure and (iii) the magnitude of maximum deviator stress if a sample of the same sand with the same void ratio were tested in a triaxial test with an all-round pressure of 60 kN/m2. Assume c = 0. (18 marks)

(i) the angle of internal friction, Direct shear box results Normal load = 0.36 kN ∴ Normal stress at failure 𝜎𝜎 = 0.36

36 × 10−4= 100 kN/m2

Similarly, 𝜏𝜏 = 0.1836 × 10−4

= 50 kN/m2 Using Coulomb’s equation 𝜏𝜏 = 𝜎𝜎 . tanф tanф = 𝜏𝜏

𝜎𝜎= 50

100= 0.50

ф = tan−1 0.50 = 26.6° But, 𝜏𝜏 is also equal to 𝜎𝜎 tanβm βm = ф = 26.6° (ii) the magnitude and direction of the principal stresses in the zone of failure Also using equation, we have 𝜎𝜎1 =

𝜎𝜎at failure1 − sin βm

𝜎𝜎1 = 100

1 − sin 26.6°= 100

1−0.448

= 181 kN/m2

Similarly, 𝜎𝜎3 = 𝜎𝜎at failure1+ sin βm

= 1001+0.448

= 69 kN/m2 The angle made by failure plane with the major principal plane is given as:

α = 45° + βm2

= 45° + 13.3°


Page 116

= 58.3° That means the failure plane makes an angle of 58.3° with the major principal plane. Also the failure plane is horizontal (as is there in a direct shear box test). Hence, the major principal plane makes an angle of 58.3° with the horizontal and the minor principal plane makes 90° with the major principal plane, i.e. 90° − 58.3° = 31.7° with the horizontal. These values and positions etc. can also be worked out graphically by Mohr circle. (iii) the magnitude of maximum deviator stress if a sample of the same sand with the same void ratio

were tested in a triaxial test with an all-round pressure of 60 kN/m2. Assume c = 0. In a triaxial testing, 𝜎𝜎3 = 60 kN/m2,σd = ? ,βm = 26.6° To find σd we must find σ1 So, using relation between σ1 and σ3, i.e. equation we have

𝜎𝜎1𝜎𝜎3

= 1+ sin βm1 − sin βm

= 1+ sin ф1− sin ф

𝜎𝜎1 = σ3(1+sin 26.6°)

1 − sin 26.6°)

= 60 × (1+0.448)1−0.448

kN/m2

= 60 × (1.448)0.552

= 157.4 kN/m2

Now, 𝜎𝜎1 = 𝜎𝜎d = 𝜎𝜎3 𝜎𝜎d = 𝜎𝜎1 − 𝜎𝜎3 = 157.4− 60 = 97.4 kN/m2 25. A 2.2 m square footing is located at a depth of 4.4 m in a stiff clay of saturated unit weight 21 kN/m3. The undrained strength of clay at a depth of 4.4 m is given by parameter w = 120 kN/m2 and 𝛗𝛗𝐮𝐮 = 0. For a factor of safety 3, with respect to shear failure, compute (i) the net value of bearing capacity (ii) the value of maximum load that could be carried by the footing. (12 marks) (i) the net value of bearing capacity Df = 4.4 m; B = 2.2 m

DfB

= 4.42.2

= 2 Since, it is a deep footing, we shall use Skempton’s equations qf = cu . Nc′ + 𝑦𝑦. Df for фμ = 0 where, cu = 120 kN/m2; 𝑦𝑦sat = 21 kN/m2 Value of Nc′ for a square footing with

DfB

= 4.42.2

= 2

Nc′ = 5 �1 + 0.2 DfB� for strip footing

Nc(s)′ = Nc′ × 1.2 for square and circular footing

= 5 �1 + 0.2 DfB� × 1.2


Page 117

= 6(1 + 0.2 × 2) = 8.4 (To be limited to 1.2 × 7.5 = 9) qf = 120 × 8.4 + 21 × 4.4 = 1100.4 kN/m2 Now, net ultimate bearing capacity qnf = ultimate bearing capacity qnf – γD = 1100.4− 21 × 4.4 = 1008 kN/m2 (ii) the value of maximum load that could be carried by the footing.

Now F = �qnf – γDfq – γDf

�

or 3 = 1100 .4−92.4𝑞𝑞 − 92.4

or 𝑞𝑞 − 92.4 = 10083

= 336

𝑞𝑞 = 336 + 92.4 = 428.4 kN/m Hence, the total load that can be taken by the footing of area 2.2 m × 2.2 m = 428.4 × 2.2 × 2.2 = 2073.5 kN

2007

26. A 12 m thick bed of sand is underlain by a layer of clay 6 m thick. The water table that was originally at ground level is lowered by drainage to a depth 4 m, whereupon the degree of saturation above lowered water reduces to 20%. Determine the increase in the effective pressure at mid of clay layer due to water table lowering. Given saturated densities of sand and clay as 2.05 g/cm3 and 1.85 g/cm3 and the dry density of sand = 1.76 g/cm3. NOTE: [g/cm3 = 103 kg/m3 x 9.8 m/s2 = 9.81 kN/m2] (20 marks) CASE 1: When water table is at ground level The effective pressure at datum A–A (i.e. middle of clay layer) w.r.t. to Fig. is given as:

σ1′ = [γsub for sand × 12 m depth + γsub for clay × 3 m depth] where, γsat for sand = ρsat . g = 2.05 x 9.81 = 20.11 kN/m3


Page 118

γsat for clay = 1.85 × 9.81 = 18.15 kN/m3 γsub for sand = 20.11 − 9.81 i.e. 𝑦𝑦w = 10.3 kN/m3 And, γsub for clay = 18.15 – 9.81 = 8.34 kN/m3 σ1′ = 10.3 × 12 + 8.34 × 3 = 148.62 kN/m2 CASE 2: When the water table gets lowered bt 4 m; then the effective stress at A–A would be given as: σ2′ = [γsand × 4 m depth + γsub for sand × 8 m depth + γsub for clay × 3 m depth] ……(i) Now, γsand for 4 m layer is to be determined, where the water content is reduced to such an extent that the degree of saturation reduces to 20% of the original. ∴ S = 0.2 ρdry = 1.76 given γdry = 1.76 × 9.81 kN/m3 = 17.27 kN/m3 γdry = 20.11 kN/m3 γ = ? γ for sand can be determined by using. γ = γdry + 𝑆𝑆(γsat − γdry ) = 17.27 + 0.2(20.11− 17.27) = 17.84 kN/m3 Substituting this value of in (i), we get σ2′ = [17.84 × 4 + 10.3 × 8 + 8.34 × 3] = 178.78 kN/m2 ∴ Increase in effective pressure σ2′ − σ1′ = 178.78− 148.62 = 30.16 kN/m2 NOTE: Lowering of water – table increases the effective stress. 27. An earth embankment is compacted at water content of 17% to a bulk density of 1.9 g/cc. If the specific gravity of soil grains is 2.65, calculate the void ratio of the compacted embankment. (10 marks) Given w = 17% = 0.17 ρ = 1.9 g/cc G = 2.65, e = ?

ρ = �(G+Se )1+e

� ρw

1.9 = �2.65+ 0.17 × 2.651+𝑒𝑒

� x 1

1 + e = 1.6318 e = 0.6318 So that the void ratio of the compacted embankment is e = 0.63

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SOIL MECHANICS & FOUNDATION ENGINEERING · SOIL MECHANICS & FOUNDATION ENGINEERING ECA Page 95 SOIL...

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