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

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