Ch10-Slope Stability Examples 2

Soil Mechanics

Assist. Prof. Dr. Rafi' M. S.

SOLVED PROBLEMS Problem (1): (Ordinary method of slices or Fellenius method) For the slope shown in the figure below, find the safety factor against sliding for the trial slip surface AC. Use the ordinary method of slices.

Problem (1): Ordinary method of slices.

Solution:

The sliding wedge is divided into 7 slices. The weight (W) of each slice and its tangential and normal components are calculated and tabulated in the following table. αn Slice W No. (kN/m) (deg.) (1) (2) (3) 1 2 3 4 5 6 7

22.4 294.4 435.2 435.2 390.4 268.8 66.58

70 54 38 24 12 0 -8

sin α n (4) 0.94 0.81 0.616 0.407 0.208 0 -0.139

cos α n (5) 0.342 0.588 0.788 0.914 0.978 1 0.990

∑❑

∆ Ln (m) (6)

W n sin α n (kN/m) (7)

2.924 6.803 5.076 4.376 4.09 4 3.232

21.1 238.5 268.1 177.1 81.2 0 -9.25

6.7 173.1 342.94 397.8 381.8 268.8 65.9

776.75 kN/m

1637.04 kN/m

30.501 m

W n cos α n (kN/m) (8)

Soil Mechanics

Assist. Prof. Dr. Rafi' M. S. col.8 ∑ ¿ tan ∅ ¿ c ( ∑ col .6 ) +¿ c L+ ∑ W cos α tan ∅ F . S .= =¿ ∑ W sin α

F . S .=¿

20 tan¿ 1.55 ¿ 20 ( 30.501 )+ ( 1637.04 ) ¿ ¿

Problem (2): (Bishop’s simplified method) Using Bishop’s simplified method, determine the short term stability of the slope shown in the figure below. Given that the slope was initially submerged with water and that the water level has now been drawn down to the level of the top of the sand.

Problem (2): Bishop’s simplified method. Solution:

Initially the centre and radius of the failure plane must be assumed. The calculations presented below are for one such assumption. However, to find the factor of safety of the slope, a number of centres and radii will be needed to find the combination that gives the minimum factor of safety. Problem calculations for slice 6: ∆ xi xi

θi

= 1.0 m measured from the figure = 2.5 m measured from the figure = sin-1 (2.5/5.83) = 25.4o or measure from the figure. Note that θ is positive for slices giving positive overturning moments

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Soil Mechanics

Assist. Prof. Dr. Rafi' M. S.

Wi = A γ = (1)(2)(15) + (1)(0.268)(20) = 35.36 kN/m Wi sin θi = 35.36 sin (25.4) = 15.17 kN/m ui = γ ω Z = (9.81)(0.268) = 2.628 kN/m ci ∆ xi + (Wi – ui ∆ xi) tan ∅ i = (0)(1) + [35.36 −¿ (2.628)(1)] tan 30o = 18.9 kN/m Note that it is ∅ the friction angle, not θ in this calculation

Now assume a factor of safety, say F = 3 = cos θ i (1 + tan θ i tan ∅ i /F) = cos(25.4) [1+ tan (25.4) tan(30)/3] = 0.986 or read Mi off the chart for θ = 25.4 and (tan ∅ /F = tan(30)/3 = 0.19

Mi

The results for all the slices can be similarly evaluated and tabulated as shown below: Slice No.

1 2 3 4 5 6 7 8

θ

∆ x

o

(m)

()

-25.4 -14.9 -4.93 4.93 14.89 25.4 36.87 50.53

F=¿

1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0

u (kPa)

2.628 6.227 7.942 7.942 6.227 2.628 -

∑ T ¿ / M =¿ ∑ W sinθ

W (kN/m)

Wsin

c

θ

∆

(kN/m)

x (kN/m )

5.357 12.70 23.69 38.69 42.70 35.36 24.96 10.62

-2.30 -3.77 -2.03 3.317 10.98 15.17 14.98 8.20

143 .3 =¿ 44 . 54

3.22

Then using the updated F=3.22 re-evaluate M and

25.0 25.0

T¿ / M

T* = c ∆ x + (W −¿ u ∆ x) tan

M

T*/M

0.821 0.917 0.980 1.013 1.016 0.986 0.800 0.636

1.92 4.08 9.28 17.52 20.73 19.17 31.26 39.30

∅ (kN/m)

1.58 3.74 9.09 17.75 21.06 18.9 25 25

until the solution converges. In this

problem this gives F = 3.25 Problem (3): (Taylor Chart c −∅ soil ) A slope has an inclination of 30 o and is 8 m high. The soil properties are c u = 20 kN/m3, ∅ u =5o, γ bulk = 15 kN/m3. Determine the short term factor of safety if the clay deposit is infinitely deep.

30o 3

8m

Soil Mechanics

Assist. Prof. Dr. Rafi' M. S.

Solution:

From Taylor stability chart, Fig.(10.18) for i = 30o and  = 5o:

F= Hence

c  011 .  HF

c 20 = =1.5 γHN s (15)(8)(0.11)

However, for the correct solution a factored tan-1[(tan /F] should be used. So having determined F.O.S. an iterative procedure is required using the updated * to determine the correct factor of safety. Regions on the chart indicate that the failure in this problem is a deep-seated failure mode.

Problem (4): (Taylor’s Stability Chart for ∅=0 Soils ) A cut 7.5m deep is to be made in a stratum of highly cohesive soil as shown in the figure below. 1. The slope angle β is 30°. 2. Bedrock is located 12m below the original ground surface. 3. The soil has a unit weight of 18.86 kN/m3; cohesion = 31.1 kN/m2 and angle of internal friction = 0°. Find the F.O.S. against slope failure.

7.5m

Cohesive soil c 31.1 kN/m2 0° 18.86 kN/m3

30o

12m

Rock

Solution:

D H = 12m and H = 7.5m, so D = 12/7.5 = 1.60 β=¿ 30o and D = 1.6;

From stability chart, Fig.(10.19) with (Taylor’s stability number)

Ns

¿

cd γHF

4

;

F=

Ns

¿

0.17

31.1 =¿ 1.29 ( 18.86 )( 7.5 ) ( 0.17)

Soil Mechanics

Assist. Prof. Dr. Rafi' M. S.

Problem (5): (Single Slice on a Planner Surface) Find the Factor of Safety of the slope cut shown in the figure below. Given that c = 7.2 kN/m2, ∅ = 25° and γ = 16.50 kN/m3. Permeable Soil:

16.5 kN/m3

1 3m

1

4.5m

30o

Highly Impermeable Cohesive Soil

Solution:

L=

D 3 = =¿ 6m sin α sin 30

h=

D sin β

w=

sin( β−α)=¿

L h γ ( 6 ) ( 1.1 ) (16.5) = =¿ 2 2

1.5m

3 sin 45

45o

sin(45−30)=¿ 1.1m

54.45 kN/m

30 25 tan ¿ ¿ ¿ ¿ 30 2.39 sin ¿ ¿ cos ¿ ¿ ( 7.2 )( 6 )+(54.45)¿ c L+W cos α tan ∅ F . S .= =¿ W sin α

Problem (6): (Culmann’s method) A 1.8 m deep vertical trench wall is to be dug in a soil without shoring. The soil’s unit weight, angel of internal friction, and cohesion are 19.0 kN/m 3, 28°, and 20.2 kN/m2 respectively. Find the Factor of safety of this trench using the Culmann method.

Note: This method assumes that sliding occurs along plane passes through the toe of the slope.

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Soil Mechanics

Assist. Prof. Dr. Rafi' M. S.

Solution:

4 c d si n β cos ∅ d

H=¿

γ [ 1−cos ⁡( β−∅ d ) ]

Try a F.S. tan ∅d =¿

tan ∅ tan28 = =¿ 0.532 F . S .∅ 1.0

Therefore, d = 28 and = 90for a vertical wall.

( 4 ) ( c d ) sin 90 cos 28

1.8=¿

F . S .c =

19 [ 1−cos(90−28) ]

; cd = 5.14 kN/m2

c 20.2 = =¿ 3.93 kN/m2 c d 5.14

Since [F.S.c ≠ [ F.S.another trial factor of safety must be attempted. Try a F.S. tan ∅d =¿

tan ∅ tan28 = =¿ 0.265 F . S .∅ 2.0

Therefore, d = 14.89 and = 90for a vertical wall.

( 4 ) ( c d ) sin 90 cos 14.89

1.8=¿

F . S .c =

19 [ 1−cos(90−14.89) ]

; cd = 6.57 kN/m2

c 20.2 = =¿ 3.07 kN/m2 c d 6.57

Since [F.S.c ≠ [ F.S.another trial factor of safety must be attempted. Try a F.S. tan ∅d =¿

tan ∅ tan28 = =¿ 0.177 F . S .∅ 3.0

Therefore, d = 10.04 and = 90for a vertical wall. 1.8=¿

( 4 ) ( c d ) sin 90 cos 10.04 19 [ 1−cos(90−10.04) ]

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; cd = 7.17 kN/m2

Soil Mechanics

Assist. Prof. Dr. Rafi' M. S. F . S .c =

c 20.2 = =¿ 2.82 kN/m2 c d 7. 17

Since [F.S.c ≠ [ F.S.the correctfactor of safety has not yet been found. Rather than continue the trial and error procedure, the values of F . S .c and F . S .∅ are plotted in figure, from which the applicable factor of safety of about 2.84 can be read.

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