module 3 soil mechanics assignment clash (final)

7/24/2019 Module 3 Soil Mechanics Assignment CLash (Final)

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UNESCO-IHE Institute for Water Education

Program: Coastal Engineering & Port Development

Module 3: Introduction to Coastal Science and Engineering

Course: Soil Mechanics

Lecturer: ir. J.H. van Dalen

Date of Submission: Monday February 15, 2016

Prepared by: Christopher H. Lashley

Student Number: 49499

Locker Number: 501


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Name: Christopher H. Lashley Student Number: 49499

ContentsForeword ................................................................................................................................................. 1

Assignment 1. .......................................................................................................................................... 2

Using the Fellenius method of slices, determine the factor against sliding for the trial slip surface. 2

Assignment 2. .......................................................................................................................................... 6

A. Determine the Active Resultant Force per unit length of the wall. ............................................ 7

B. Determine the Passive Resultant force per unit length of the wall. ........................................... 8

C. Assuming a strut at surface level, calculate the safety factor against failure of the passive soil.

Also draw the conclusion if the soil will collapse or not. .................................................................... 9

D. Motivate (no calculation) how answer c) would be effected if the water table on both sides

would be deeper, with D=0............................................................................................................... 10

E. Motivate (no calculation) how answer c) would be effected if the strut level is chosen deeper.

10

References ............................................................................................................................................ 14


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1

Foreword

This document includes a short written report in response to Assignments 1 and 2 provided. It includes

the answers to the questions posed, as well as a short explanation of the steps taken and sample

calculations to show the theory applied to solve each problem. Computer programs, specifically

Google Sketch Up and Microsoft Excel, were used to accurately represent and analyse the problems

posed. As mentioned, sample calculations have also been provided to support the tables and diagrams

produced. As instructed, the following values (see bold text) were used based on the last digit of my

student number (49499):

Assignment Parameter Last digit from your student number:

0 1 2 3 4 5 6 7 8 9

1 Xm[m] 2 2.5 3 3.5 4 4.5 5 5 5 5.5

Ym[m] 8 8.5 9 9.5 10 10.5 11 11.5 12 12

Yc[m] -2.5 -2 -2.5 -2 -2.5 -2 -1.5 -1 -1.5 -1

2 D [m] 6 3.5 5 2.5 4 3.5 3 5.5 2 1.5


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Assignment 1.

Using the Fellenius method of slices, determine the factor against sliding for the trial

slip surface.

General Approach:

1. The slip circle was determined by the coordinates Xm= 5.5 mand Ym= 12 mof the centre

point and the deepest level of circle Yc= -1 m(values based on student number). This was

drawn using the graphical program, Google SketchUp (see Figure 1 below).

2.

The failure body was divided into 8 slices, as recommended. The boundaries of the slices were

selected based on the locations of discontinuity in stratigraphy and geometry of the soil

profile. The main areas of concern were: the change in slope and the change in soil properties

between layers. This resulted in slices of varied widths, ranging from 1.64 m to 2.67 m.

3.

For each slice, the dimensions bi and hi, angle i (obtained graphically), weight Wi, shear

strength parameters at the slip surface c iand Iwere determined.

4. Finally, the stability factor was calculated according to Fellenius by determining the ratio of

the Resisting Moment to the Driving Moment.

Figure 1 Slip Circle to be examined based on Xm= 5.5 m, Ym= 12 m and Yc= -1 m, Google SketchUp


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Soil Properties:

Soil Unit weight, [kN/m3] Friction angle, ' [o] Cohesion, c [kPa]

1 17 32 0

2 17 17.5 10

Sample Calculations:

For slice No. 1,

Figure 2 Slice No. 1

Graphically: h1 = 1.33 m, b1= 1.69 m and 1= 55.4o

Slip surface in Soil 1, therefore: '1= 32oand c1= 0

Unit weight (Same for soil 1 and 2) = 17 kN/m 3

Effective vertical stress,

1.33 17 / 22.61 /

Weight,

22.61 / 1.69 m 38.21 /

Length of slip surface,

1.69

55.4 2.98


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Contribution of Slice No. 1 to Resisting Moment,

+

0 2.98 + 38.21

55.4 32 13.56 /

Contribution of Slice No.1 to Driving Moment,

38.21 / 55.4 31.45 /

The above parameters were determined for each slice. Finally the Safety Factor against sliding for the

trial slip surface was calculated using the following expression:

(= + )

=


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Results:

The results of the analysis are provided in the table below:

Slice No. hi(m) bi(m) ni(kN/m2) Wi(kN/m) i(o) sini cosi li(m) DMi (kN/m) Wi cosai (kN/m) 'i (o) tan'i c' (kPa) RMi (kN/m)

1 1.33 1.69 22.61 38.21 55.4 0.82 0.57 2.98 31.45 21.70 32 0.62 0 13.56

2 3.35 1.64 56.95 93.40 43.9 0.69 0.72 2.28 64.76 67.30 32 0.62 0 42.05

3 4.63 1.64 78.71 129.08 34.6 0.57 0.82 1.99 73.30 106.25 17.5 0.32 10 53.43

4 5.14 2.67 87.38 233.30 23.6 0.40 0.92 2.91 93.40 213.79 17.5 0.32 10 96.55

5 4.67 2.67 79.39 211.97 11.2 0.19 0.98 2.72 41.17 207.93 17.5 0.32 10 92.78

6 3.69 2.67 62.73 167.49 -0.7 -0.01 1.00 2.67 -2.05 167.48 17.5 0.32 10 79.51

7 2.23 1.82 37.91 69.00 -10.8 -0.19 0.98 1.85 -12.93 67.77 17.5 0.32 10 39.90

8 0.83 1.82 14.11 25.68 -19.1 -0.33 0.94 1.93 -8.40 24.27 17.5 0.32 10 26.91

280.71 444.68

444.68280.71

.


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Assignment 2.

The following diagram of the retaining wall and soil profile was produced using Google SketchUp. The

location of the water table was determined by taking D = 1.5 m (based on the student number 49499):

Figure 3 Diagram of Retaining Wall showing location of Water Table and Soil Profile

Soil and Water Properties:

Soil Dry Unit weight,

[kN/m3]

Saturated Unit weight,

sat[kN/m3]

Friction angle, ' [o] Cohesion, c [kPa]

Sand 18 20 32 0

Clay 16 17 7

Kindly note the following general considerations used for the analysis:

Unit weight of water, w= 9.81 kN/m3.

In the active case, no tension crack zone develops in Sand Layer (c = 0)(Ishibashi & Hazarika,

2015).

In the passive case, there is a positive pressure at z = 0 (c 0) and there is no tension crack

zone (Ishibashi & Hazarika, 2015).


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A. Determine the Active Resultant Force per unit length of the wall.

General Approach:

1. The effective vertical stress was calculated at depths, z = 0, 5.5, 7.0 and 14.0 m by calculating

the total vertical stress and subtracting the hydrostatic water pressure.

2.

The coefficients of active horizontal soil pressure (Ka) were then determined for the Sand and

Clay layers.

3. The active horizontal soil pressure, awas then calculated at depths, z = 0, 5.5, 7.0 and 14.0

m and the resulting pressure distribution was determined and divided into various

components (see Figure).

4. The hydrostatic water pressure acting on the active side of the wall, z*wwas also calculated

at depths, z= 7.0 and 14.0 m and the resulting pressure distribution was determined andanalysed as one of the components of the overall active force.

5. This Active Force, Ac was then determined by finding the summation of the areas of the

individual components of the active horizontal pressure distribution (Aci), including the

hydrostatic water pressure.


At depth z = 5.5 m,

Effective vertical stress(equal to total vertical stress since water table is below point of interest),

5.5 18 99 /

Coefficient of active horizontal soil pressure (Sand layer),

1 1 +

1 321 + 32

0.3073

Active horizontal pressure,

2 0.307 99 0.307 2 0 30.42 /

Contribution to Active Force,

12

12

30.42 5.5 83.66 /

The Total Active Resultant Force per unit length of the wall was found to be: 935.74 kN/m.


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B. Determine the Passive Resultant force per unit length of the wall.

General Approach:

1. The effective vertical stress was calculated at depths, z = 7.0 and 14.0 m by calculating the

total vertical stress and subtracting the hydrostatic pressure.

2.

The coefficients of passive horizontal soil pressure (Kp) were then determined for the Clay

layer.

3. The passive horizontal soil pressure, pwas then calculated at depths, z = 7.0 and 14.0 m and

the resulting pressure distribution was determined and divided into various components.

4.

The hydrostatic pressure acting on the passive side of the wall, hw*wwas also calculated at

depths, z= 7.0 and 14.0 m and the resulting pressure distribution was determined and

analysed as one of the components of the overall passive force.5. This Resultant Passive Force, Pa was then determined by finding the summation of the areas

of the individual components of the passive horizontal pressure distribution (Pai).


At depth z = 14.0 m,

Effective vertical stress,

( ) + ( ) ( ) (7.0 16) + (1.5 9.81) (8.5 9.81) 43.33 /

Coefficient of passive horizontal soil pressure (Clay Layer),

1+1

1 + 171 17

1.826

Passive horizontal pressure,

+ 2

1.826 43.33 + 1.826 2 7 98.06 /

Contribution to Passive Force (Soil),

18.92 7.0 132.44 /

12

12

(98.06 18.92) 7.0 276.97 /

Please Note: subscripts refer to component number (see Passive Pressure Distribution Diagram).

The Total Passive Resultant Force per unit length of the wall was found to be: 763.80 kN/m.


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C. Assuming a strut at surface level, calculate the safety factor against failure of the

passive soil. Also draw the conclusion if the soil will collapse or not.

General approach:

1.

Moments, Maiand Mpiwere taken about the strut location by multiplying the individual

force components, Aci and Paiby their lever arms, li; with liequal to the perpendicular

distance from the line of action for each force component (this was found to bez for

triangular components andz for rectangular components of the pressure distribution

diagram) to the location of the strut.

2. The Factor of Safety was then calculated by dividing the sum of the Resisting (Passive)

Moments, Mpiby the sum of the Driving (Active) Moments, Mai.


Driving Moment for Active Force Component #1,

83.66 3.67 307.03 /

Resisting Moment for Passive Force Component #8,

276.97 11.67 3231.36 /

Safety Factor,

8579.29 /9378.78 /

0.91

Therefore, the soil will collapse.


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D. Motivate no calculation) how answer c) would be effected if the water table on

both sides would be deeper, with D=0.

As the water table drops, the effective vertical and horizontal stresses on the active side of the wall

will increase. At D=0 the passive side will remain submerged and there will be no change in effective

vertical or horizontal stresses on the passive side. The net effect of this will be to increase the active

force on the wall resulting in a lower factor of safety.

Figure 4 Increase in effective stress in sand layer when D=0

E.

Motivate no calculation) how answer c) would be effected if the strut level is

chosen deeper.

As the strut level was originally taken at the surface, all forces on the active side produced a

clockwise moment and all forces on the passive side an anti-clockwise moment. However, should

the strut level be chosen much deeper (say 4 m), the soil above that level on the active side would

now create an anti-clockwise moment (passive soil state) and add to the total passive moment,

and reduce the total active moment, . This would result in an increase in the safety

factor.


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Figure 5 New failure mode as a result of lower strut level

Results:

Soil Type ' (deg) dry(kN/m3) sat(kN/m3) c' (kPa) Ka Kp

Sand 32 18 20 0 0.3073 3.255

Clay 17 16 7 0.548 1.826

Active pressure

Soil

Type

Depth, z

(m)

H

(m)

n

(kN/m2)

hw*w (kN/m2) 'n

(kN/m2)

c'

(kN/m2)

'a

(kN/m2)

0.00 0.00 0.00 0.00 0.00 0.00 0.00

Sand 5.50 5.50 99.00 0.00 99.00 0.00 30.42

Sand 7.00 1.50 129.00 14.72 114.29 0.00 35.12

Clay 7.00 1.50 129.00 14.72 114.29 7.00 52.22

Clay 14.00 7.00 241.00 83.39 157.62 7.00 75.94

Passive Pressure

Soil

Type

Depth, z

(m)

H

(m)

n

(kN/m2)

hw*w (kN/m2) 'n

(kN/m2)

c'

(kN/m2)

'p

(kN/m2)

0.00 0.00 0.00 0.00 0.00 0.00 0.00

5.50 5.50 0.00 0.00 0.00 0.00 0.00

Clay 7.00 1.50 14.72 14.72 0.00 7.00 18.92

Clay 14.00 7.00 126.72 83.39 43.33 7.00 98.06

New failure mode due to rotation about lower strut level

Passive States


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Active Force

Component No. Ac (kN/m) l (m) Ma (kNm/m)

1 83.65 3.67 306.72

2 45.63 6.25 285.17

3 3.52 6.50 22.904 365.51 10.50 3837.91

5 83.04 11.67 968.77

6 354.39 11.17 3957.31

935.74 9378.78

Passive Force

Component No. Pa (kN/m) l (m) Mp (kNm/m)

7 132.44 10.50 1390.61

8 276.97 11.67 3231.36

9 354.39 11.17 3957.31

763.80 8579.29

Safety Factor 0.91

Figure 6 Active Pressure Distribution and Force Components


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Figure 7 Passive Pressure Distribution and Force Components

Figure 8 Hydrostatic Pressure Distribution and Force Components


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References

Craig, R. F., 1992. Soil Mechanics. 5th ed. s.l.:Chapman & Hall.

Ishibashi, I. & Hazarika, H., 2015. Soil Mechanics Fundamentals and Applications. s.l.:CRC Press.

Verruijt, A., 2001. Soil Mechanics. Delft: Delft University of Technology.

module 3 soil mechanics assignment clash (final)

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