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Class notes of Geotechnical Engineering course I used to teach at UET Lahore. Feel free to download the slide show.

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- 1. 1 Geotechnical Engineering–II [CE-321] BSc Civil Engineering – 5th Semester by Dr. Muhammad Irfan Assistant Professor Civil Engg. Dept. – UET Lahore Email: mirfan1@msn.com Lecture Handouts: https://groups.google.com/d/forum/geotech-ii_2015session Lecture # 24 6-Dec-2017
- 2. 2 Practice Problem #4 sin1 sin1 2 45tan2 o a aK aaa KczK 2 c’ = 0 ’ = 35° = 18 kN/m3 c’ = 0 ’ = 30° WT = 19 kN/m3 sat = 21 kN/m3 c’ = 0 ’ = 32° sat = 20 kN/m3 4m 4m 2m 4m Determine the total active force per meter acting on the wall along with its point of application. q = 50 kPa
- 3. 3 Practice Problem #5 sin1 sin1 2 45tan2 o a aK aaa KczK 2 c’ = 10 kPa ’ = 35° = 18 kN/m3 WT c’ = 50 ’ = 0° sat = 20 kN/m3 4m 4m 4m Determine the total active force per meter acting on the wall along with its point of application. q = 50 kPa c’ = 20 kPa ’ = 19.5° sat = 21 kN/m3
- 4. 4 Practice Problem #6 sin1 sin1 2 45tan2 o a aK aaa KczK 2 c’ = 50 kPa ’ = 10° = 18 kN/m3 10 m A retaining wall of 10 m height retains a cohesive soil. Determine the active force with respect to various possibilities of tension crack. a c K c z 2
- 5. 5 RANKINE THEORY ACTIVE PRESSURE -- SUMMARY -- sin1 sin1 2 45tan2 o a aK aaa KczK 2 a c K c z 2
- 6. 6 COULOMB’S EARTH PRESSURE THEORY ASSUMPTIONS 1. The soil is homogeneous and isotropic. 2. Soil has both cohesion and friction (c- soil). 3. Rupture surface as well as backfill surface is planar. 4. There is friction between wall and soil. 5. Failure wedge is a rigid body undergoing translation. Coulomb (1776)
- 7. 7 BENEFITS OF ASSUMPTIONS -- DIFFERENCE BETWEEN THEORY AND REALITY --
- 8. 8 BENEFITS OF ASSUMPTIONS -- DIFFERENCE BETWEEN THEORY AND REALITY -- Theoretical Earth Pressure Actual Earth Pressure
- 9. 9 COULOMB’S ACTIVE EARTH PRESSURE a b q 180aq ab A B C D qb W W = Weight of soil wedge ABC 𝑊 = 1 2∙𝐴𝐶∙𝐵𝐷∙1∙𝛾 ⋯⋯⋯(1) 𝐴𝐶 sin(𝛼 + 𝛽) = 𝐴𝐵 sin(𝜃 − 𝛽) ∆𝑨𝑩𝑪 Using law of sines 𝐴𝐶 = 𝐴𝐵 sin(𝜃 − 𝛽) ∙ sin(𝛼 + 𝛽) 𝐴𝐶 = 𝐻 sin 𝛼 ∙ sin(𝜃 − 𝛽) ∙ sin(𝛼 + 𝛽) H
- 10. 10 COULOMB’S ACTIVE EARTH PRESSURE a b q 180aq ab A B C D qb W W = Weight of soil wedge ABC 𝑊 = 1 2∙𝐴𝐶∙𝐵𝐷∙1∙𝛾 ⋯⋯⋯(1) 𝐴𝐶 = 𝐻 sin 𝛼 ∙ sin(𝜃 − 𝛽) ∙ sin(𝛼 + 𝛽) ∆𝑨𝑩𝑫 𝐵𝐷 sin(180 − (𝛼 + 𝜃)) = 𝐴𝐵 sin 90 ∵ sin(180 − 𝛼 + 𝜃 ) = sin(𝛼 + 𝜃) 𝐵𝐷 sin(𝛼 + 𝜃) = 𝐴𝐵 sin 90 𝐵𝐷 = sin(𝛼 + 𝜃) ∙ 𝐴𝐵 1 𝐵𝐷 = sin(𝛼 + 𝜃) ∙ 𝐻 sin 𝛼 H
- 11. 11 COULOMB’S ACTIVE EARTH PRESSURE a b q 180aq ab A B C D qb W W = Weight of soil wedge ABC 𝑊 = 1 2∙𝐴𝐶∙𝐵𝐷∙1∙𝛾 ⋯⋯⋯(1) 𝐴𝐶 = 𝐻 sin 𝛼 ∙ sin(𝜃 − 𝛽) ∙ sin(𝛼 + 𝛽) 𝐵𝐷 = sin(𝛼 + 𝜃) ∙ 𝐻 sin 𝛼 𝑊 = 1 2 ∙ 𝛾𝐻2 𝑠𝑖𝑛2 𝛼 ∙ sin(𝛼 + 𝛽) ∙ sin(𝛼 + 𝜃) sin(𝜃 − 𝛽) Eq. 1 → H
- 12. 12 COULOMB’S ACTIVE EARTH PRESSURE a b q 180aq ab A B C D qb W W R R = Resultant of shear and normal forces acting on failure plane R (q) (ad)180(adq) d Pa 𝛿 = 2 3 𝜙 Our Goal: Determine active force (Pa) on the wall. Draw force polygon of the system. Pa d = angle of wall friction (𝑅𝑒𝑠𝑜𝑛𝑎𝑏𝑙𝑒 𝑎𝑠𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛)
- 13. 13 COULOMB’S ACTIVE EARTH PRESSURE a b q 180aq ab A B C D qb W W R R (q) (ad)180(adq) d Pa 𝑃𝑎 sin(𝜃 − 𝜙) = 𝑊 sin[180 − 𝛼 − 𝛿 + 𝜃 − 𝜙 ] Applying sine law on force polygon Pa 𝑃𝑎 sin(𝜃 − 𝜙) = 𝑊 sin(𝛼 − 𝛿 + 𝜃 − 𝜙) Replacing value of ‘W’ 𝑃𝑎 = 1 2 ∙ 𝛾𝐻2 𝑠𝑖𝑛2 𝛼 ∙ sin(𝛼 + 𝛽) ∙ sin(𝛼 + 𝜃) ∙ sin(𝜃 − 𝜙) sin(𝜃 − 𝛽) ∙ sin(𝛼 − 𝛿 + 𝜃 − 𝜙)
- 14. 14 COULOMB’S ACTIVE EARTH PRESSURE a b q 180aq ab A B C D qb W W R R (q) (ad)180(adq) d Pa As designers, we want to determine max. value of Pa Pa 𝑃𝑎 = 1 2 𝛾𝐻2 ∙ 𝑠𝑖𝑛2(𝛼 + 𝜙) 𝑠𝑖𝑛2 𝛼 ∙ sin(𝛼 − 𝛿) 1 + sin(𝜙 + 𝛿) ∙ sin(𝜙 − 𝛽) sin(𝜙 − 𝛿) ∙ sin(𝜙 + 𝛽) 2 To determine critical value of b for max. Pa, we have 𝑑𝑃𝑎 𝑑𝛽 = 0
- 15. 15 COULOMB’S ACTIVE EARTH PRESSURE a b q 180aq ab A B C D qb W R d Pa 𝐾 𝑎 = 𝑠𝑖𝑛2(𝛼 + 𝜙) 𝑠𝑖𝑛2 𝛼 ∙ sin(𝛼 − 𝛿) 1 + sin(𝜙 + 𝛿) ∙ sin(𝜙 − 𝛽) sin(𝜙 − 𝛿) ∙ sin(𝜙 + 𝛽) 2 Since, 𝑃𝑎 = 1 2 ∙ 𝛾𝐻2 ∙ 𝐾 𝑎 𝑃𝑎 = 1 2 𝛾𝐻2 ∙ 𝑠𝑖𝑛2 (𝛼 + 𝜙) 𝑠𝑖𝑛2 𝛼 ∙ sin(𝛼 − 𝛿) 1 + sin(𝜙 + 𝛿) ∙ sin(𝜙 − 𝛽) sin(𝜙 − 𝛿) ∙ sin(𝜙 + 𝛽) 2
- 16. 16 COULOMB’S ACTIVE EARTH PRESSURE a b q 180aq ab A B C D qb W R d Pa 𝑃𝑎 = 1 2 𝛾𝐻2 ∙ 𝑠𝑖𝑛2 (𝛼 + 𝜙) 𝑠𝑖𝑛2 𝛼 ∙ sin(𝛼 − 𝛿) 1 + sin(𝜙 + 𝛿) ∙ sin(𝜙 − 𝛽) sin(𝜙 − 𝛿) ∙ sin(𝜙 + 𝛽) 2 For a vertical wall face and horizontal levelled ground 𝛼 = 90° , 𝑎𝑛𝑑 𝛽 = 0° 𝑃𝑎 = 1 2 𝛾𝐻2 ∙ 1 − sin 𝜙 1 + sin 𝜙 Above equation is reduced to i.e. same as Renkine’s Solution
- 17. 17 CONCLUDED REFERENCE MATERIAL Principles of Geotechnical Engineering – (7th Edition) Braja M. Das Chapter #13 Essentials of Soil Mechanics and Foundations (7th Edition) David F. McCarthy Chapter #17 Geotechnical Engineering – Principles and Practices – (2nd Edition) Coduto, Yueng, and Kitch Chapter #17

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