For the concrete Dam shown in figure, determine the total average seepage for the given permeable strata by drawing Flow Net. The detail dimensions are given on the figure. The head of water on upstream side is 6ft and on downstream side is 1ft. Assume of Nf= 3, Nd= 6 and rectangular flow elements.

 

Q#:For the concrete Dam shown in figure, determine the total average seepage for the given permeable strata by drawing Flow Net. The detail dimensions are given on the figure. The head of water on upstream side is 6ft and on downstream side is 1ft. Assume of Nf= 3, Nd= 6 and rectangular flow elements. 

Solution#

 ANSWER 

 SOLUTION

For rectangular channel

n=b/L = 2/4 =0.5

seepage equation for rectangular chanel                                                                           

 

                                                                                                                 k = 4×10^-4 ft/day

q = kH (Nf / Nd)                                                                                       H = h₁ - h₂

                                                                                                                 H = 6-1 = 5

No. of potetial drop = 6                                                                           Nf =  3

No. of Flow lines     = 3                                                                           Nd = 6

H = 6 - 1 = 5'

 

Rate of Seepage

 

 q = nf / nd kh

  

q = 3/6×4×10^-4 × 5

 q = 1×10^-3 ft³ /day/ft

The following are the results of different tests on soil samples are provided in the following table. Soil sample Percent passing sieve Liquid Limit plasticity Index No 10 No 40 No 200 A 100 75 25 35 15 B 100 85 30 45 22 C 80 45 10 25 12 Classify the soil according to AASTHO classification and determine the Group Index in each case. You may use the information provided.

 

Q#3 :The following are the results of different tests on soil samples are provided in the following table.                      

Soil sample	Percent passing sieve			Liquid Limit	plasticity Index
	No 10	No 40	No 200
A	100	75	25	35	15
B	100	85	30	45	22
C	80	45	10	25	12

Classify the soil according to AASTHO classification and determine the Group Index in each case. You may use the information provided.

ANSWER

SOIL Sample  A

      Percentage passing through  No. 10  (2mm) seive = 100%

      Percntage passing  through No. 40 (425 μm) seive = 75%

      Percentage passing through No. 200 (75 μm) seive = 25%

 

              Liquid limit = wL = 35

             Plasticity index = I_p = 15

Percenrage passing through no. 200 seieve  < 35%

        So there for soil is granual

because Percentage passing through no.10 (2mm) seive > 50%

Percentage passing through no. 40 (425 μm) seive > 50%

Percentage passing through no.200(75 μm) >10%  < 35%

There fore soil group is A -2

 Because  wL < 40    and    I_p > 11

 

There fore soil subgroup is A-2-6

 Group index of the soil is given as

GI = 0.2a = 0.005ac +0.01 bd

a = percentage passing through  75 μm seive  -35                     a  ≤  40

b = percentage passing through 75 μm seive  -15                       b ≤  40

c = wL -40        c  ≤  20

d = I_p– 10        d  ≤  20

 

Therefore,

                        a= 25-35= -10

                        b =25-15 =10

                        c = 35-40 = -5

                        d = 15-10 = 5

GI= 0.2 × (-10)+0.005 ×(-10)(-5)+0.01 × (10)(5)

GI = -2+0.25+0.5

GI = -1.5

There is no negative group index of soil is zero

There for soil classified is  A-2-6

 

SOIL Sample  B

      Percentage passing through  No. 10  (2mm) seive = 100%

      Percntage passing  through No. 40   (425 μm) seive = 85%

      Percentage passing through No. 200  (75 μm) seive = 30%    (My reg no.)

 

              Liquid limit = wL = 45

             Plasticity index = I_p = 22

 

Percenrage passing through no. 200 seieve  >35%

        So there for soil is silt clay type

because wL  > 41   and I_p   >  11 

 

There fore soil group is A -7

(wL  - 30) = 45 -30 =15

 Because  and    I_p > (wL  - 30)

There fore soil subgroup is A-7-6

 Group index of the soil is given as

GI = 0.2a = 0.005ac +0.01 bd

a = percentage passing through  75 μm seive  -35                     a  ≤  40

b = percentage passing through 75 μm seive  -15                       b ≤  40

c = wL -40        c  ≤  20

d = I_p– 10        d  ≤  20

 

Therefore,

                        a= 30-35=-5

                        b =30-15 =15

                    

                        c = 45-40 = 5

                        d = 22-10 = 12

GI= 0.2 × (-5) +0.005 × (-5)  ×5 +0.01 × 15 ×12

GI = 0.675

The group index of soil is = 0.675

There for soil classified is  A-7-6(0.675)        (clay soil)

 

SOIL Sample  C

      Percentage passing through  No. 10  (2mm) seive = 80%

      Percntage passing  through No. 40 (425 μm) seive = 45%

      Percentage passing through No. 75 (75 μm) seive = 10%

 

              Liquid limit = wL = 25

             Plasticity index = I_p = 12

 

Percentage passing through no. 200 seieve  < 35%

        So there for soil is granual

because Percentage passing through no.10 (2mm) seive > 50%

Percentage passing through no. 40 (425 μm) seive < 50%

Percentage passing through no.200(75 μm) >10%  < 25%

 

There fore soil subgroup is A-2-6

Group index of the soil is given as

GI = 0.2a = 0.005ac +0.01 bd

a = percentage passing through  75 μm seive  -35                     a  ≤  40

b = percentage passing through 75 μm seive  -15                       b ≤  40

c = wL -40        c  ≤  20

d = I_p– 10        d  ≤  20

 

Therefore,

                        a= 10-35= - 25

                        b =10-15 = -5

                        c = 25-40 = 15

                        d = 12-10 = 2

 

GI= 0.2 × (-25)+0.005 ×(-25)(-15)+0.01 × (-5)(2)

GI = -5+1.875+0.01

GI = -3.135

There for group index of soil is zero

There for soil classified is  A-1-b(0)

Derive and explain Darcy’s Law in light of its application in design of foundation for the discharge of water through the soil sample.

Q#:Derive and explain Darcy’s Law in light of its application in design of foundation for the discharge of water through the soil sample.     

ANSWER

Darcy’s law;

    Darcy’s law states the principle which governs the movement of fluid in the given substance. Darcy’s law equation that describes the capability of the liquid to flow via any porous media like a rock. The law is based on the fact according to which, the flow between two points is directly proportional to the pressure differences between the points, the distance, and the connectivity of flow within rocks between the points. Measuring the inter-connectivity is known as permeability. 

Darcy's law demonstrat experimently that for leminar flow condition in a saturated soil mass,  the velocity of flow is directly proportional to hydraulic gradients

 i.e  V ∝ i

Or   V = ki

Where K is coefficient of permeability

   I is  hydraulic gradient

I =  Δh/ L

Δ h = change in head b.w two observation points.

L= length b.w the two points

Now considering applipcation in the design of foundation the equation for discharge can be derived  from a flow net as

 

Δ q =  k× Δh/ L

Δh = H/Nd where Nd is number of potential drops 

Δ q=  Kh/ L × 1/Nd ×b

Or

Δ q= kH/Nd × (b/L)

 

The total discharge  through flow net will be

I= h₁ -h₂ /L

Or

v=k (h₁  -h₂ ) / L

 Now discharge of water through the soil sample

q=v × area= v×A

or

q= k (h₁-h₂) / L ×L

 

q = Nf × Δ q

= Nf × k H/Nd × b/L

q = kH Nf/Nd × b/L

The field is approximate square b= L

 q = kH Nf/Nd

 

or

Total head = P/w + v²/2g +z

 piezometric head = p/w +z

 In soil mechanics  v²/2g is neglected as  it is very small

 

Flow occurs only when there is difference  in total head b.w 2point 

Total head difference is      H₁ -H₂ =  h_f

                                                          h_f= H_{1 -}H₂ 

 

Total head loss b.w 2 poits in a soil is  equal to the difference in the elevation of water level in the 2 piezometer kept @ those 2 poits

 

Seepage length = L

hydraulic gradient = h/L  (Head loss per unit length)

c/s of section x-x

c/s area of soil  = A = b×d

Mutual area of flow = Av = nA (    AREA of voids)

 discharge velocity/apparent velocity of flow = Q/A

 seepage velocity/ actual velocity Vs = Q/Av

 

Q= AV =Av ×Vs

Vs = A/Av  (v)

 

     Vs = v/n                Vs > V

Derive and explain Darcy’s Law in light of its application in design of foundation for the discharge of water through the soil sample.

Derive and explain Darcy’s Law in light of its application in design of foundation for the discharge of water through the soil sample.       

Solution

ANSWER

Darcy’s law;

    Darcy’s law states the principle which governs the movement of fluid in the given substance. Darcy’s law equation that describes the capability of the liquid to flow via any porous media like a rock. The law is based on the fact according to which, the flow between two points is directly proportional to the pressure differences between the points, the distance, and the connectivity of flow within rocks between the points. Measuring the inter-connectivity is known as permeability.

 

Darcy's law demonstrat experimently that for leminar flow condition in a saturated soil mass,  the velocity of flow is directly proportional to hydraulic gradients

 i.e  V ∝ i

Or   V = ki

 

Where K is coefficient of permeability

   I is  hydraulic gradient

I =  Δh/ L

Δ h = change in head b.w two observation points.

L= length b.w the two points

Now considering applipcation in the design of foundation the equation for discharge can be derived  from a flow net as

Δ q =  k× Δh/ L

Δh = H/Nd where Nd is number of potential drops

Δ q=  Kh/ L × 1/Nd ×b

Or

Δ q= kH/Nd × (b/L)

The total discharge  through flow net will be

I= h₁ -h₂ /L

Or

v=k (h₁  -h₂ ) / L

 Now discharge of water through the soil sample

q=v × area= v×A

or

q= k (h₁-h₂) / L ×L

 

q = Nf × Δ q

= Nf × k H/Nd × b/L

q = kH Nf/Nd × b/L

 

The field is approximate square b= L

 

 q = kH Nf/Nd

or

Total head = P/w + v²/2g +z

 piezometric head = p/w +z

 

 In soil mechanics  v²/2g is neglected as  it is very small

Flow occurs only when there is difference  in total head b.w 2point

Total head difference is      H₁ -H₂ =  h_f

                                                          h_f= H_{1 -}H₂ 

Total head loss b.w 2 poits in a soil is  equal to the difference in the elevation of water level in the 2 piezometer kept @ those 2 poits

 

Seepage length = L

hydraulic gradient = h/L  (Head loss per unit length)

 

c/s of section x-x

c/s area of soil  = A = b×d

Mutual area of flow = Av = nA (    AREA of voids)

 

 discharge velocity/apparent velocity of flow = Q/A

 seepage velocity/ actual velocity Vs = Q/Av

 

Q= AV =Av ×Vs

Vs = A/Av  (v)

 

     Vs = v/n                Vs > V

 

Provide three applications of each in light of soil stability. I. Hydraulic conductivity II. Liquid limit III. Plasticity index IV. Moisture content V. Permeability

 

Q#:Provide three applications of each in light of soil stability.

                                            I.            Hydraulic conductivity                                                 

                                         II.            Liquid limit

                                       III.            Plasticity index

                                      IV.            Moisture content

                                         V.            Permeability

ANSWER

 

Hydraulic coductivity

 

Ø More the hydraulic conductivity  of soil,more  will be the movement of water inside soil thus reducing stability of soil

Ø Hydraulic conductivity value is a measure of compactio e.g high hydraulic conductivity value indicate that the soil is not well compacted . so we can infer that as a weak soi 

Ø Increased hydraulic conductivity of water can lead to accumulation of pores pressureinside soil which in turn decrease the effective stress acting on soil thus  create probability of bulking of soil.

 

plastic index

Ø higher plasticity index represent more  plastic behaviour of soil which means more settlement of soil,thus reducing soil stability.

Ø soil with low or zero plasticity index have stiff or non plastic nature which show high bearig capacity  of soil which increse  soil stability

Ø it indicate compressibility of soil e.g the greater the plasticity index higher the soil compressibility .

.

Moisture Content

 

Ø Soil which low moisture  content show high bearing capacity of soil thus increasing soil capacity 

Ø moisture contet effect the shear strength of soil (mainly if the soil conain clay particals ) generally shear stregth of soil decrease with increase in water.

Ø Moisture soil contain value effect the penetration resistance value e.g dry soil have high penetration resistance and vice versa.

 

liquid limits

Ø liquid limit values can be used to predict the consolidation properties and state of consistency of fine grained soil on site

Ø liquid limit values are used to measure the allowable bearing capacity

Ø liquid limit values are used to measure the settlement value due to any loading o top of soil

Permeability

Ø permeability values gives an idea about the soil texture which directly related to soil stability

Ø Higher permeability means presence of large voids which results in increased settlment thus reducing stability of soil and It gives an idea about soil structure or packing of molecule which in tur relate to soil stability

Ø high permeable soil are more susceptible to soil slope failure due to increase movement of water inside soil