module 7 (lecture 25) retaining walls topics -...

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Module 7

(Lecture 25)

RETAINING WALLS

Topics

Check for Bearing Capacity Failure Example Factor of Safety Against Overturning Factor of Safety Against Sliding Factor of Safety Against Bearing Capacity Failure

1.6 OTHER TYPES OF POSSIBLE RETAINING WALL FAILURE

Check for Bearing Capacity Failure

The vertical pressure as transmitted to the soil by the base slab of the retaining wall should be checked against the ultimate bearing capacity of the soil. The nature of variation of the vertical pressure transmitted by the base slab into the soil is shown in figure 7.10. Note that qtoe and qheel are the maximum and the minimum pressures occurring at the ends of the toe and heel sections, respectively. The magnitudes of qtoe and qheel can be determined in the following manner.


Figure 7.10 Check for bearing capacity failure

The sum of the vertical forces acting on the base slab is Σ 𝑉𝑉 (see column 3, table 2), and the horizontal force is 𝑃𝑃𝑎𝑎 cos𝛼𝛼. Let R be the resultant force, or

→𝑅𝑅 = →

Σ 𝑉𝑉 +→

(𝑃𝑃𝑎𝑎 cos 𝛼𝛼) [7.15]

The net moment of these forces about point C (figure 7.10) is

𝑀𝑀net = Σ 𝑀𝑀𝑅𝑅 − Σ 𝑀𝑀𝑂𝑂 [7.16]

Note that the values of Σ 𝑀𝑀𝑅𝑅 and Σ 𝑀𝑀𝑂𝑂 have been previously determined (see column 5, table 2 and equation (3)]. Let the line of action of the resultant, R, interest the base slab at E, as shown in figure 7.10. The distance 𝐶𝐶𝐶𝐶 then is

𝐶𝐶𝐶𝐶�� = 𝑋𝑋� = 𝑀𝑀netΣ 𝑉𝑉

[7.17]

Hence the eccentricity of the resultant, R, may be expressed as

𝑒𝑒 = 𝐵𝐵2− 𝐶𝐶𝐶𝐶�� [7.18]

The pressure distribution under the base slab may be determined by using the simple principles of mechanics of materials:


𝑞𝑞 = Σ 𝑉𝑉𝐴𝐴

± 𝑀𝑀net 𝑦𝑦𝐼𝐼

[7.19]

Where

𝑀𝑀net = moment = (Σ 𝑉𝑉)𝑒𝑒

𝐼𝐼 = moment of inertia per unit length of the base section = 112

(1)(𝐵𝐵2)

For maximum and minimum pressures, the value of y in equation (19) equals 𝐵𝐵/2. Substituting the preceding values into equation (19) gives

𝑞𝑞max = 𝑞𝑞toe = Σ 𝑉𝑉(𝐵𝐵)(1)

+𝑒𝑒(Σ 𝑉𝑉)𝐵𝐵2� 1

12�(𝐵𝐵2)= Σ 𝑉𝑉

𝐵𝐵�1 + 6𝑒𝑒

𝐵𝐵� [7.20]

Similarly,

𝑞𝑞min = 𝑞𝑞heel = Σ 𝑉𝑉𝐵𝐵�1 − 6𝑒𝑒

𝐵𝐵� [7.21]

Note that Σ 𝑉𝑉 includes the soil weight, as shown in table 2, and that, when the value of the eccentricity, e, becomes greater than 𝐵𝐵/6, 𝑞𝑞min becomes negative [equation (21)]. Thus, there will be some tensile stress at the end of the heel section. This stress is not desirable because the tensile strength of soil is very small. If the analysis of a design shows that 𝑒𝑒 > 𝐵𝐵/6, the design should be reproportioned and calculations redone.

The relationships for the ultimate bearing capacity of a shallow foundation were discussed in chapter 3. Recall that

𝑞𝑞𝑢𝑢 = 𝑐𝑐2𝑁𝑁𝑐𝑐𝐹𝐹𝑐𝑐𝑐𝑐𝐹𝐹𝑐𝑐𝑐𝑐 + 𝑞𝑞𝑁𝑁𝑞𝑞𝐹𝐹𝑞𝑞𝑐𝑐𝐹𝐹𝑞𝑞𝑐𝑐 + 12𝛾𝛾2𝐵𝐵′𝑁𝑁𝛾𝛾𝐹𝐹𝛾𝛾𝑐𝑐𝐹𝐹𝛾𝛾𝑐𝑐 [7.22]

Where

𝑞𝑞 = 𝛾𝛾2𝐷𝐷

𝐵𝐵′ = 𝐵𝐵 − 2𝑒𝑒

𝐹𝐹𝑐𝑐𝑐𝑐 = 1 + 0.4 𝐷𝐷𝐵𝐵′

𝐹𝐹𝑞𝑞𝑐𝑐 = 1 + 2 tan𝜙𝜙2(1 − sin𝜙𝜙2)2 𝐷𝐷𝐵𝐵′

𝐹𝐹𝛾𝛾𝑐𝑐 = 1

𝐹𝐹𝑐𝑐𝑐𝑐 = 𝐹𝐹𝑞𝑞𝑐𝑐 = �1 − 𝜓𝜓 °

90°�2

𝐹𝐹𝛾𝛾𝑐𝑐 = �1 − 𝜓𝜓 °

𝜙𝜙2°�

2


𝜓𝜓° = tan−1 �𝑃𝑃𝑎𝑎 cos 𝛼𝛼Σ 𝑉𝑉

�

Note that the shape factors 𝐹𝐹𝑐𝑐𝑐𝑐 ,𝐹𝐹𝑞𝑞𝑐𝑐 , and 𝐹𝐹𝛾𝛾𝑐𝑐 given in chapter 3 are equal to 1 because they can be treated as a continuous foundation. For this reason, the shape factors are not shown in equation (22).

Once the ultimate bearing capacity of the soil has been calculated by using equation (22), the factor of safety against bearing capacity failure can be determined.

𝐹𝐹𝐹𝐹(bearing capacity ) = 𝑞𝑞𝑢𝑢𝑞𝑞max

[7.23]

Generally, a factor of safety of 3 is required. In chapter 3 we noted that the ultimate bearing capacity of shallow foundations occurs at a settlement of about 10% of the foundation width. In the case of retaining walls, the width B is large. Hence the ultimate load 𝑞𝑞𝑢𝑢 will occur at a fairly large foundation settlement. A factor of safety of 3 against bearing capacity failure may not ensure, in all cases, that settlement of the structure will be within the tolerable limit. Thus this situation needs further investigation.

Example 1

The cross section of a cantilever retaining wall is shown in figure 7.11. Calculate the factors of safety with respect ot overturning and sliding and bearing capacity.

Figure 7.11


Solution

Referring to figure 7.11,

𝐻𝐻′ = 𝐻𝐻1 + 𝐻𝐻2 + 𝐻𝐻3 = 2.6 tan 10° + 6 + 0.7 = 0.458 + 6 + 0.7 = 7.158 m

The Rankine active force per unit length of wall= 𝑃𝑃𝑝𝑝 = 12𝛾𝛾1𝐻𝐻′2𝐾𝐾𝑎𝑎 . For 𝜙𝜙1 = 30°,𝛼𝛼 = 10°,𝐾𝐾𝑎𝑎 is

equal to 0.350 (table 2 from chapter 6). Thus,

𝑃𝑃𝑎𝑎 = 12(18)(7.158)2(0.35) = 161.4 kN/m

𝑃𝑃𝑣𝑣 = 𝑃𝑃𝑎𝑎 sin 10° = 161.4(sin 10°) = 28.03 kN/m

𝑃𝑃ℎ = 𝑃𝑃𝑎𝑎 cos 10° = 161.4(cos 10°) = 158.95 kN/m

Factor of Safety Against Overturning

The following table can now be prepared for determination of the resisting moment:

Section no. Area (𝑚𝑚2) Weight/unit length (kN/m)

Moment from C (kN/m)

Moment (kN − m)

1 6 × 0.5 = 3 70.74 1.15 81.35

2 12(0.2)6 = 0.6 14.15 0.833 11.79

3 4 × 0.7 = 2.8 66.02 2.0 132.04

4 6 × 2.6 = 15.6 280.80 2.7 758.16

5 12(2.6)(0.458)

= 0.595

10.71 3.13 33.52

𝑃𝑃𝑣𝑣 =28.03 4.0 112.12

Σ𝑉𝑉 = 470.45 Σ 1128.98

= Σ 𝑀𝑀𝑅𝑅

For section numbers, refer to figure 7.11,

𝛾𝛾concrete = 2358 kN/m3


The overturning moment, 𝑀𝑀𝑂𝑂

𝑀𝑀𝑂𝑂 = 𝑃𝑃ℎ �𝐻𝐻′3� = 158.95 �7.158

3� = 379.25 kN − m

𝐹𝐹𝐹𝐹(overturning ) = Σ𝑀𝑀𝑅𝑅𝑀𝑀𝑂𝑂

= 1128.98379.25

= 2.98 > 2 − OK

Factor of Safety Against Sliding

From equation (11)

𝐹𝐹𝐹𝐹(sliding ) =(Σ 𝑉𝑉) tan (𝑘𝑘1𝜙𝜙2)+𝐵𝐵𝑘𝑘2𝑐𝑐2+𝑃𝑃𝑝𝑝

𝑃𝑃𝑎𝑎 cos 𝛼𝛼

Let 𝑘𝑘1 = 𝑘𝑘2 = 23

Also

𝑃𝑃𝑝𝑝 = 12𝐾𝐾𝑝𝑝𝛾𝛾2𝐷𝐷2 + 2𝑐𝑐2�𝐾𝐾𝑝𝑝𝐷𝐷

𝐾𝐾𝑝𝑝 = tan2 �45 + 𝜙𝜙22

= tan2(45 + 10) = 2.04�

𝐷𝐷 = 1.5 m

So

𝑃𝑃𝑝𝑝 = 12(2.04)(19)(1.5)2 + 2(40)(�2/04) (1.5)

= 43.61 + 171.39 = 215 kN/m

Hence

𝐹𝐹𝐹𝐹(sliding ) =(470.45)tan �2×20

3 �+(4)�23�(40)+215

158.95

= 111.5+106.67+215158.95

= 2.73 > 1.5 − OK

Note: For some designs, the depth D for passive pressure calculation may be taken to be equal to the thickness of the base slab.

Factor of Safety Against Bearing Capacity Failure

Combining equations (16, 17 and 18),

𝑒𝑒 = 𝐵𝐵2− Σ 𝑀𝑀𝑅𝑅−Σ 𝑀𝑀𝑂𝑂

Σ 𝑉𝑉= 4

2− 1128.98−379.25

470.45

= 0.406m < 𝐵𝐵6

= 46

= 0.666 m


Again, from equations (20 and 21)

𝑞𝑞heeltoe = Σ 𝑉𝑉

𝐵𝐵�1 ± 6𝑒𝑒

𝐵𝐵� = 470.45

4�1 ± 6×0.406

4� =

189.2 kN/m2 (toe)45.9 kN/m2(heel)

The ultimate bearing capacity of the soil can be determined from equation (22):

𝑞𝑞𝑢𝑢 = 𝑐𝑐2𝑁𝑁𝑐𝑐𝐹𝐹𝑐𝑐𝑐𝑐𝐹𝐹𝑐𝑐𝑐𝑐 + 𝑞𝑞𝑁𝑁𝑞𝑞𝐹𝐹𝑞𝑞𝑐𝑐𝐹𝐹𝑞𝑞𝑐𝑐 + 12𝛾𝛾2𝐵𝐵′𝑁𝑁𝛾𝛾𝐹𝐹𝛾𝛾𝑐𝑐𝐹𝐹𝛾𝛾𝑐𝑐

For 𝜙𝜙2 = 20° (table 4 from chapter 3), 𝑁𝑁𝑐𝑐 = 14.83,𝑁𝑁𝑞𝑞 = 6.4 and 𝑁𝑁𝛾𝛾 = 5.39. Also

𝑞𝑞 = 𝛾𝛾2𝐷𝐷 = (19)(1.5) = 28.5 kN/m2

𝐵𝐵′ = 𝐵𝐵 − 2𝑒𝑒 = 4 − 2(0.406) = 3.188 m

𝐹𝐹𝑐𝑐𝑐𝑐 = 1 + 0.4 �𝐷𝐷𝐵𝐵′� = 1 + 0.4 � 1.5

3.188� = 1.188

𝐹𝐹𝑞𝑞𝑐𝑐 = 1 + 2 tan𝜙𝜙2(1 − sin𝜙𝜙2)2 �𝐷𝐷𝐵𝐵′� = 1 + 0.315 � 1.5

3.188� = 1.148

𝐹𝐹𝛾𝛾𝑐𝑐 = 1

𝐹𝐹𝑐𝑐𝑐𝑐 = 𝐹𝐹𝑞𝑞𝑐𝑐 = �1 − 𝜓𝜓 °

90°�2

𝜓𝜓 = tan−1 �𝑃𝑃𝑎𝑎 cos 𝛼𝛼Σ 𝑉𝑉

� = tan−1 �158.95470.45

� = 18.67°

So

𝐹𝐹𝑐𝑐𝑐𝑐 = 𝐹𝐹𝑞𝑞𝑐𝑐 = �1 − 18.67°

90� = 0.628

𝐹𝐹𝛾𝛾𝑐𝑐 = �1 − 𝜓𝜓𝜙𝜙�

2= �1 − 18.67

20� ≈ 0

Hence

𝑞𝑞𝑢𝑢 = (40)(14.83)(1.188)(0.628) + (28.5)(6.4)(1.148)(0.628) + 12(19)(5.93)(3.188)(1)(0)

= 442.57 + 131.50 + 0 = 574.07 kN/m2

𝐹𝐹𝐹𝐹(bearing capacity ) = 𝑞𝑞𝑢𝑢𝑞𝑞toe

= 574.07189.2

= 3.03 > 3 − OK

Example 2

A concrete gravity retaining wall is shown in figure 7.12. Determine

a. The factor of safety against overturning


b. The factor of safety against sliding c. The pressure on the soil at the toe and heel

Figure 7.12

(Note: Unit weight of concrete = 𝛾𝛾𝑐𝑐 = 150 lb/ft3).

Solution

𝐻𝐻′ = 15 + 2.5 = 17.5 ft

𝐾𝐾𝑎𝑎 = tan2 �45 − 𝜙𝜙12� = tan2 �45 − 30

2� = 1

3

𝑃𝑃𝑎𝑎 = 12𝛾𝛾(𝐻𝐻′)2𝐾𝐾𝑎𝑎 = 1

2(121)(17.5)2�13� = 6176 lb/ft

= 6.716 kip/ft

Since 𝛼𝛼 = 0

𝑃𝑃ℎ = 𝑃𝑃𝑎𝑎 = 6.176 kip/ft

𝑃𝑃𝑣𝑣 = 0

Part a: Factor of Safety Against Overturning


The following table can now be prepared to obtain Σ 𝑀𝑀𝑅𝑅:

Area (from figure 7. 12)

Weight (kip) Moment arm from C(ft) Moment about C (kip/ft)

1 12(0.8)(15)(𝛾𝛾𝑐𝑐)

= 0.9 1.25 + 2

3(0.8) = 1.783 1.605

2 (1.5)(15)(𝛾𝛾𝑐𝑐)= 3.375

1.25 + 0.8 + 0.75 = 2.8 9.45

3 12(5.25)(15)(𝛾𝛾𝑐𝑐)

= 5.906 1.25 + 0.8 + 1.5 +

5.253

= 5.3

31.30

4 (10.3)(2.5)(𝛾𝛾𝑐𝑐)= 3.863

10.32

= 5.15 19.89

5 12(5.25)(15)(0.121)

= 4.764 1.25 + 0.8 + 1.5

+ 23(5.25)

= 7.05

33.59

6 (1.5)(15)(0.121)= 2.723

21.531

1.25 + 0.8 + 1.5 + 5.25+ 0.75= 9.55

26.0

121.84 = 𝑀𝑀𝑅𝑅

The overturning moment

𝑀𝑀𝑂𝑂 = 𝐻𝐻′3𝑃𝑃𝑎𝑎 = �17.5

3� (6.176) = 36.03 kip/ft

𝐹𝐹𝐹𝐹(overturning ) = 121.8436.03

= 3.38

Part b: Factor of Safety Against Sliding

From equation (11), with 𝑘𝑘1 = 𝑘𝑘2 = 23 and assuming that 𝑃𝑃𝑝𝑝 = 0,

𝐹𝐹𝐹𝐹(sliding ) =Σ 𝑉𝑉 tan �2

3�𝜙𝜙2+𝐵𝐵�23�𝑐𝑐2

𝑃𝑃𝑎𝑎


=21.531 tan �2×20

3 �+10.3�23�(1.0)

6.176

= 5.1+6.876.176

= 1.94

Part c: Pressure on the Soil at the Toe and Heel

From equations (16, 17 and 18),


Σ 𝑉𝑉= 10.3

2− 121.84−36.03

21.531= 5.15 − 3.99 = 1.16 ft

𝑞𝑞toe = Σ VB�1 + 6e

𝐵𝐵� = 21.531

10.3�1 + (6)(1.16)

10.3� = 3.5 kip/ft2

𝑞𝑞heel = Σ VB�1 − 6e

𝐵𝐵� = 21.531

10.3�1 − (6)(1.16)

10.3� = 0.678 kip/ft2

Example 3

Repeat example 2 and use Coulomb’s active pressure for calculation and 𝛿𝛿 = 2𝜙𝜙/3.

Solution

Refer to figure 7.13 for the pressure calculation:

Figure 7.13

𝛿𝛿 = 23𝜙𝜙 = �2

3�(30) = 20°

From table 5 (chapter 6), 𝐾𝐾𝑎𝑎 = 0.4794(𝛼𝛼 = 0°,𝛽𝛽 = 70°), so


𝑃𝑃𝑎𝑎 = 12(0.121)(17.5)2(0.4794) = 8.882 kip/ft

𝑃𝑃ℎ = 𝑃𝑃𝑎𝑎 cos 40 = (8.882)(cos 40) = 6.8 kip/ft

𝑃𝑃𝑣𝑣 = 𝑃𝑃𝑎𝑎 sin 40 = 5.71 kip/ft

Part a: Factor of Safety Against Overturning

Refer to figure 7. 14 and 12.

Area (from figure 7. 12 and 14)

Weight (kip) Moment arm from C(ft) Moment about C (kip/ft)

1 0.9𝑎𝑎 1.783𝑎𝑎 1.605

2 3.375𝑎𝑎 2.8𝑎𝑎 9.46

3 5.906𝑎𝑎 5.3𝑎𝑎 31.30

4 3.863𝑎𝑎 5.15𝑎𝑎 19.89

𝑃𝑃𝑣𝑣 = 5.71

19.75

1.25 + 0.8 + 1.5 + 5.25− 121= 7.59

43.34

105.6

Same as in example 2

The overturning moment is

𝑀𝑀𝑂𝑂 = 𝑃𝑃ℎ𝐻𝐻′3

= (6.8) �17.53� = 39.67 kip/ft

Hence

𝐹𝐹𝐹𝐹(overturning ) = 105.639.67

= 2.66


Figure 7.14

Part b: Factor of Safety Against Sliding

𝐹𝐹𝐹𝐹(sliding ) =Σ 𝑉𝑉 tan �2

3�𝜙𝜙2+𝐵𝐵�23�𝑐𝑐2

𝑃𝑃ℎ

=19.75 tan �2

3�(20)+10.3�23�(1.0)

6.8= 1.7

Part c: Pressure on the Soil at the Toe and Heel


Σ 𝑉𝑉= 10.3

2− (105.6−39.67)

19.67= 1.8 ft

𝑞𝑞toe = 19.7510.3

�1 + (6)(1.8)10.3

� = 3.93 kip/ft2

𝑞𝑞heel = 19.7510.3

�1 − (6)(1.8)10.3

� = −0.093 kip/ft2 ≈ 0

OTHER TYPES OF POSSIBLE RETAINING WALL FAILURE

In addition to the three types of possible failure for retaining walls discussed in section 4, two other types of failure could occur: shallow shear failure and deep shear failure.

Shallow shear failure in soil below the base of a retaining wall takes place along a cylindrical surface 𝑎𝑎𝑎𝑎𝑐𝑐 passing through the heel, as shown in figure 7.15a. The center of the arc of the circle 𝑎𝑎𝑎𝑎𝑐𝑐 is located at 𝑂𝑂, which is found by trial and error (corresponds to the minimum factor of safety). This type of failure can occur as the result of excessive induced shear stress along the


cylindrical surface in soil. In general, the factor of safety against horizontal sliding is lower than the factor of safety obtained by shallow shear failure, if 𝐹𝐹𝐹𝐹(sliding ) is greater than about 1.5, shallow shear failure under the base may not occur.

Figure 7.15 (a) Shallow shear failure; (b) deep shear failure

Deep shear failure can occur along a cylindrical surface 𝑎𝑎𝑎𝑎𝑐𝑐 as shown in figure 7.15b, as the result of the existence of a weak layer of soil underneath the wall at a depth of about 1.5 times the width of the retaining wall. In such cases, the critical cylindrical failure surface 𝑎𝑎𝑎𝑎𝑐𝑐 has to be determined by trial and error with various centers, such as O (figure 7.15b). The failure surface along which the minimum factor of safety is obtained is the critical surface of sliding. For the backfill slope with 𝛼𝛼 less than about 10°, the critical failure circle apparently passes through the edge of the heel slab (such as 𝑐𝑐𝑒𝑒𝑑𝑑 in figure 7.15b). In this situation, the minimum factor of safety also has to be determined by trial and error by changing the center of the trial circle.

The following is an approximate procedure for determining the factor of safety against deep-seated shear failure for a gently sloping backfill (𝛼𝛼 < 10°) developed by Teng (1962). Refer to 7.16.


Figure 7.16 Deep shear failure analysis

1. Draw the retaining wall and the underlying soil layer to a convenient scale. 2. For a trial center O, draw an arc of a circle 𝑎𝑎𝑎𝑎𝑐𝑐𝑐𝑐. For all practical purposes, the weight of

the soil in the area 𝑎𝑎𝑎𝑎𝑐𝑐𝑐𝑐𝑒𝑒 is symmetrical about the vertical line drawn through point O. let the radius of the trial circle be r.

3. To determine the driving force on the failure surface causing instability (figure 7.16a), divide the area in the zone 𝑒𝑒𝑑𝑑𝑒𝑒ℎ into several slices. These slices can be treated as rectangles or triangles, as the case may be.

4. Determine the area of each of these slices and then determine the weight W of the soil (and/or concrete) contained inside each slice (per unit length of the wall).

5. Draw a vertical line through the centroid of each slice, and locate the point of intersection of each vertical line with the trial failure circle.

6. Join point O (that is, the center of the trial circles) with the points of intersection as determined in step 5.

7. Determine the angle, 𝜔𝜔, that each vertical line makes with the radial line.


8. Calculate 𝑊𝑊 sin𝜔𝜔 for each slice. 9. Determine the active force 𝑃𝑃𝑎𝑎 on the face 𝑐𝑐𝑑𝑑, 1

2𝛾𝛾1𝐻𝐻′2𝐾𝐾𝑎𝑎 . 10. Calculate the total driving force:

Σ (𝑊𝑊 sin𝜔𝜔) + 𝑃𝑃𝑎𝑎𝑋𝑋�

𝑟𝑟 [7.24]

Where 𝑋𝑋� = perpendicular distance between the line of action of 𝑃𝑃𝑎𝑎 and the center 𝑂𝑂

11. To determine the resisting force on the failure surface (figure 7.16b), divide the area in the zones 𝑎𝑎𝑎𝑎𝑘𝑘 and 𝑐𝑐𝑐𝑐𝑒𝑒𝑑𝑑𝑖𝑖 into several slices, and determine the weight of each slice, 𝑊𝑊1 (per unit length of the wall). Note that points 𝑎𝑎 and 𝑐𝑐 are on top of the soft clay layer; the weight of each slice shown in figure 7.16b is 𝑊𝑊1 is contrast to the weight of each slice W, as shown in figure 7.16a.

12. Draw a vertical line through the centroid of each slice and locate the point of intersection of each line with the trial failure circle.

13. Join point O with the points of intersection as determined in step 12. Determine the angles, 𝜔𝜔1, that the vertical lines make with the radial lines.

14. For each slice, obtain 𝑊𝑊1 tan𝜙𝜙2 cos𝜔𝜔1

15. Calculate 𝑐𝑐2𝑙𝑙2 + 𝑐𝑐3𝑙𝑙2 + 𝑐𝑐2𝑙𝑙3 Where 𝑙𝑙1, 𝑙𝑙2, and 𝑙𝑙3 are the lengths of the arcs 𝑎𝑎𝑎𝑎, 𝑎𝑎𝑐𝑐, and 𝑐𝑐𝑐𝑐

16. The maximum resisting force that can be derived along the failure surface is Σ (𝑊𝑊1 tan𝜙𝜙2 cos𝜔𝜔1) + 𝑐𝑐2𝑙𝑙2 + 𝑐𝑐3𝑙𝑙2 + 𝑐𝑐2𝑙𝑙3 [7.25]

17. Determine the factor of safety against deep shear failure for this trial failure surface: 𝐹𝐹𝐹𝐹(deep shear failure ) = Σ(𝑊𝑊1 tan 𝜙𝜙2 cos 𝜔𝜔1)+𝑐𝑐2𝑙𝑙2+𝑐𝑐3𝑙𝑙2+𝑐𝑐2𝑙𝑙3

Σ (𝑊𝑊 sin 𝜔𝜔)+𝑃𝑃𝑎𝑎𝑋𝑋�𝑟𝑟

[7.26]

Several other trial failure surfaces may be drawn, and the factor of safety can be determined in a similar manner. The lowest value of the factor for safety obtained from all trial surfaces is the desired factor of safety.

module 7 (lecture 25) retaining walls topics -...

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