Sunday, October 16, 2022

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 = Ip = 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    Ip > 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 = Ip– 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 = Ip = 22

 

Percenrage passing through no. 200 seieve  >35%

        So there for soil is silt clay type

because wL  > 41   and Ip   >  11 

 

There fore soil group is A -7

(wL  - 30) = 45 -30 =15

 Because  and    Ip > (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 = Ip– 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 = Ip = 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 = Ip– 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 =  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 × Î” 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 =  hf

                                                          hf= H1 -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

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