HOMEWORK SOLUTION

 

HW #10

 

CE 353

GEOTECHNICAL ENGINEERING I

 

(Question no. 8.9, 8.10, 8.17, 8.19 ..Al-Khafaji textbook)

 

 


8.9              Undisturbed soil sample D = 50 cm tested in triaxial cell where H = 100 mm. Sheared under additional load Fd = 725 N with vertical deformation d = 18 cm failure plane was inclined at 51 degrees to the horizontal. Cell pressure s3 = 300 kPa.

 

Question:

Determine Coulomb’s equation for the soil shear strength in terms of total stress.

 

Solution:

 

a = 51o                        a = 45o+(f/2)                          f = (51-45)x2 = 12o


 

 

 

So, the equation is        …………..ans

 

 

 

8.10          Box area = 4500 mm2

Normal load = 625 N

 

Question:

Failure shear stress.

 

Solution:

 

……………….use f in problem no. 8.9

 

 


8.17

 

Sample

s3(kN/m2)

s1(kN/m2)

uo(kN/m2)

s3’

s1’

1

196.1

343.2

137.3

38.8

205.9

2

392.3

686.5

275.6

116.7

410.9

 

Question:

 

(a)    values c and f for total stresses

(b)   values c’ and f’ for effective stresses

(c)    shear and normal stress on the failure plane in sample 2

 

Solution:

 

(a)  fT = 13o           from the graph

     

c = (c1+c2)/2=(8.416.7)/2=12.6kN/m2

 

 

(b)  f’ = 25o           from the graph

 

 

c = (c1+c2)/2=(35 +39.1)/2=37kN/m2

 

 

(c) 

 

     

     

 

     

 

     

 

     

     

 

           

 

 

 

8.19     Depth   = 10 m

            LL        = 65

            PL        = 30

            gsat        = 17.8 kN/m3

 

            Question:

            Estimate Su (Undrained Shear Strength).

 

            Solution:

           

            Su = Cu = Po (0.11 + 0.0037 PI)

 

            Po = 10 x (gsat-gw) = 10 x (17.8-9.81) = 79.9 kN/m2 = 79.9 kPa

           

            PI = LL – PL = 65 – 30 = 35

 

            Su=Cu = 79.9 ( 0.11 + 0.0037 PI )

                        = 79.9 ( 0.11 + 0.0037 x 35 )

                        = 19.1 kN/m2 = 19.1 kPa