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

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/m^2)	s1(kN/m2)	uo(kN/m^2)	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) f^T = 13^o from the graph

c = (c₁+c₂)/2=(8.416.7)/2=12.6kN/m²

(b) f’ = 25^o from the graph

c = (c₁+c₂)/2=(35 +39.1)/2=37kN/m²

(c)

8.19 Depth = 10 m

LL = 65

PL = 30

g_sat = 17.8 kN/m³

Question:

Estimate Su (Undrained Shear Strength).

Solution:

Su = Cu = Po (0.11 + 0.0037 PI)

Po = 10 x (g_sat-g_w) = 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