Geotechnical Engineering-II [Lec #25: Coulomb EP Theory - Numericals]

Geotechnical Engineering-II [Lec #25: Coulomb EP Theory - Numericals]

• 1.
  1 Geotechnical Engineering–II [CE-321] BScCivil Engineering – 5th Semester by Dr. Muhammad Irfan Assistant Professor Civil Engg. Dept. – UET Lahore Email: [email protected] Lecture Handouts: https://groups.google.com/d/forum/geotech-ii_2015session Lecture # 25 8-Dec-2017
• 2.
  2 COULOMB’S ACTIVE EARTHPRESSURE a b q 180-a-q a+b A B C D q-b 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
• 3.
  3 COULOMB’S ACTIVE EARTHPRESSURE a b q 180-a-q a+b A B C D q-b W W f R = Resultant of shear and normal forces acting on failure plane R (q-f) (a-d)180-(a-d+q-f) 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)
• 4.
  4 COULOMB’S ACTIVE EARTHPRESSURE a b q 180-a-q a+b A B C D q-b W W f R = Resultant of shear and normal forces acting on failure plane R (q-f) (a-d)180-(a-d+q-f) 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)
• 5.
  5 COULOMB’S ACTIVE EARTHPRESSURE a b q 180-a-q a+b A B C D q-b W W R (q-f) (a-d)180-(a-d+q-f) 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) f
• 6.
  6 COULOMB’S ACTIVE EARTHPRESSURE a b q 180-a-q a+b A B C D q-b W W R (q-f) (a-d)180-(a-d+q-f) 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(𝛼 − 𝛿) ∙ sin(𝛼 + 𝛽) 2 To determine critical value of b for max. Pa, we have 𝑑𝑃𝑎 𝑑𝛽 = 0 𝑃𝑎 = 1 2 ∙ 𝛾𝐻2 𝑠𝑖𝑛2 𝛼 ∙ sin(𝛼 + 𝛽) ∙ sin(𝛼 + 𝜃) ∙ sin(𝜃 − 𝜙) sin(𝜃 − 𝛽) ∙ sin(𝛼 − 𝛿 + 𝜃 − 𝜙) f
• 7.
  7 COULOMB’S ACTIVE EARTHPRESSURE a b q 180-a-q a+b A B C D q-b W d Pa 𝐾 𝑎 = 𝑠𝑖𝑛2(𝛼 + 𝜑) 𝑠𝑖𝑛2 𝛼 ∙ sin(𝛼 − 𝛿) 1 + sin(𝜙 + 𝛿) ∙ sin(𝜙 − 𝛽) sin(𝛼 − 𝛿) ∙ sin(𝛼 + 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(𝛼 − 𝛿) ∙ sin(𝛼 + 𝛽) 2 f
• 8.
  8 COULOMB’S ACTIVE EARTHPRESSURE a b q 180-a-q a+b A B C D q-b W d Pa 𝑃𝑎 = 1 2 𝛾𝐻2 ∙ 𝑠𝑖𝑛2 (𝛼 + 𝜙) 𝑠𝑖𝑛2 𝛼 ∙ sin(𝛼 − 𝛿) 1 + sin(𝜙 + 𝛿) ∙ sin(𝜙 − 𝛽) sin(𝛼 − 𝛿) ∙ sin(𝛼 + 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 f
• 9.
  9 COULOMB’S PASSIVE EARTHPRESSURE Failure wedge and forces acting on it for passive earth pressure Force triangle to establish Pp 𝑃𝑝 = 1 2 𝛾𝐻2 ∙ 𝑠𝑖𝑛2(𝛼 − 𝜙) 𝑠𝑖𝑛2 𝛼 ∙ sin(𝛼 + 𝛿) 1 − sin(𝜙 + 𝛿) ∙ sin(𝜙 + 𝛽) sin(𝛼 + 𝛿) ∙ sin(𝛼 + 𝛽) 2
• 10.
  10 Practice Problem #7 f’= 35°  = 18.5 kN/m3 d = (2/3) f6 m Using Coulomb’s theory, find out Pa and its point of application. a = 90° b = 20° 𝑃𝑎 = 1 2 𝛾𝐻2 ∙ 𝑠𝑖𝑛2 (𝛼 + 𝜙) 𝑠𝑖𝑛2 𝛼 ∙ sin(𝛼 − 𝛿) 1 + sin(𝜙 + 𝛿) ∙ sin(𝜙 − 𝛽) sin(𝛼 − 𝛿) ∙ sin(𝛼 + 𝛽) 2
• 11.
  11 Practice Problem #8 Acounterfort retaining wall of 10 m height retains sand. The void ratio and angle of internal friction of the sand respectively are 0.70 and 30° in loose state and 0.40 and 40° in dense state. Compute and compare active and passive earth pressure in both cases. Take Gs = 2.66
• 12.
  12 Practice Problem #9 Avertical wall 10 m high retains sand having specific gravity of 2.65 and void ratio of 0.54. The WT is 4 m below the ground surface. The surface of the backfill is horizontal and carries a uniform surcharge of 25 kPa. The sand above the WT has an average degree of saturation of 70% and a f-value of 35°. The sand below the WT has a f-value of 30°. Determine the active thrust on the back of wall and its point of application.
• 13.
  13 Practice Problem #10 Determinethe magnitude and location of Pa for several possible alternatives that are produced due to tension crack. c’ = 20 kPa f’ = 10°  = 17.5 kN/m3 4 m f’ = 15°  = 20.5 kN/m3 2 m WT f ff   + -        -    sin1 sin1 2 45tan2 o a aK aaa KczK - 2 a c K c z    2
• 14.
  14 Practice Problem #11 f ff   + -       -    sin1 sin1 2 45tan2 o a aK aaa KczK - 2 c’ = 13.5 kPa  = 18 kN/m3 WT f’ = 30° sat = 20 kN/m3 2m 6m For a retaining wall shown in figure below, compute; a. Total active earth pressure per meter width of the wall b. Location of active thrust from the base of wall c. The total thrust behind the wall if WT is lowered to the base of wall q = 30 kPa CLAY SAND
• 15.
  23 CONCLUDED REFERENCE MATERIAL Principles ofGeotechnical 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