A Find: Ko γT=117.7 [lb/ft3] σh=2,083 Water Sand 0.36 0.45 0.54 0.63 wc=12% Find the at-rest lateral earth pressure coefficient. [pause] In this problem, γT=117.7 [lb/ft3] lb ft2 σh=2,083 @ A

A Find: Ko γT=117.7 [lb/ft3] σh=2,083 Water Sand 0.36 0.45 0.54 0.63 wc=12% a 20 foot thick layer of sand is submerged --- γT=117.7 [lb/ft3] lb ft2 σh=2,083 @ A

A Find: Ko γT=117.7 [lb/ft3] σh=2,083 Water Sand 0.36 0.45 0.54 0.63 wc=12% beneath a body of water, 10 feet deep. γT=117.7 [lb/ft3] lb ft2 σh=2,083 @ A

A Find: Ko γT=117.7 [lb/ft3] σh=2,083 Water Sand 0.36 0.45 0.54 0.63 wc=12% We’ve been given some properties of the sand layer, γT=117.7 [lb/ft3] lb ft2 σh=2,083 @ A

A Find: Ko γT=117.7 [lb/ft3] σh=2,083 Water Sand 0.36 0.45 0.54 0.63 wc=12% Noting, that point A is 15 feet beneath the soil surface. [pause] We begin by --- γT=117.7 [lb/ft3] lb ft2 σh=2,083 @ A

A Find: Ko γT=117.7 [lb/ft3] σh=2,083 Water Sand Ko= @ A lb ft2 wc=12% defining the at-rest lateral earth pressure coefficient as --- γT=117.7 [lb/ft3] lb ft2 σh=2,083 @ A

A Find: Ko σ’h σ’v γT=117.7 [lb/ft3] σh=2,083 Water Sand Ko= @ A lb wc=12% the horizontal effective stress divided by the vertical effective stress. [pause] γT=117.7 [lb/ft3] lb ft2 σh=2,083 @ A

A Find: Ko σ’h σ’v γT=117.7 [lb/ft3] σh=2,083 Water Sand Ko= @ A lb wc=12% Let’s first solve for the horizontal effective stress, at point A. γT=117.7 [lb/ft3] lb ft2 σh=2,083 @ A

A Find: Ko σ’h σ’v σ’h = σh - u γT=117.7 [lb/ft3] σh=2,083 Water Sand wc=12% The horizontal effective stress equals the total horizontal stress minus the pore water pressure. γT=117.7 [lb/ft3] lb ft2 σh=2,083 @ A

A Find: Ko σ’h σ’v σ’h = σh - u γT=117.7 [lb/ft3] σh=2,083 Water Sand wc=12% For point A, we’ve been given the total horizontal stress, which is 2,083 pounds per square feet. γT=117.7 [lb/ft3] lb σh=2,083 @ A ft2

A Find: Ko σ’h σ’v σ’h = σh - u γT=117.7 [lb/ft3] σh=2,083 u = γW * d Water 10 [ft] σ’h = σh - u Sand u = γW * d 15 [ft] 20 [ft] A wc=12% The pore water pressure is the unit weight of water times the depth of water, γT=117.7 [lb/ft3] lb σh=2,083 @ A ft2

A Find: Ko σ’h σ’v σ’h = σh - u γT=117.7 [lb/ft3] σh=2,083 u = γW * d Water 10 [ft] σ’h = σh - u Sand u = γW * d 15 [ft] 20 [ft] lb = 62.4 * (25 [ft]) A wc=12% Or, 62.4 pounds per cubic feet times 25 feet. ft3 γT=117.7 [lb/ft3] lb σh=2,083 @ A ft2

+ A Find: Ko σ’h σ’v σ’h = σh - u γT=117.7 [lb/ft3] σh=2,083 Water 10 [ft] σ’h = σh - u + Sand u = γW * d 15 [ft] 20 [ft] lb = 62.4 * (25 [ft]) A wc=12% where the 25 feet is the 10 feet of water depth plus 15 feet of soil depth. ft3 γT=117.7 [lb/ft3] lb σh=2,083 @ A ft2

+ A Find: Ko σ’h σ’v σ’h = σh - u γT=117.7 [lb/ft3] σh=2,083 Water 10 [ft] σ’h = σh - u + Sand u = γW * d 15 [ft] 20 [ft] lb = 62.4 * (25 [ft]) A wc=12% The pore water pressure is 1,560 pounds per square feet, ft3 γT=117.7 [lb/ft3] lb = 1,560 ft2 lb σh=2,083 @ A ft2

+ A Find: Ko σ’h σ’v σ’h = σh - u γT=117.7 [lb/ft3] σh=2,083 Water 10 [ft] σ’h = σh - u + Sand u = γW * d 15 [ft] 20 [ft] lb = 62.4 * (25 [ft]) A wc=12% which makes the effective horizontal stress --- ft3 γT=117.7 [lb/ft3] lb = 1,560 ft2 lb σh=2,083 @ A ft2

+ A Find: Ko σ’h σ’v σ’h = σh - u γT=117.7 [lb/ft3] σh=2,083 σ’h = 523 Water 10 [ft] σ’h = σh - u + Sand u = γW * d 15 [ft] 20 [ft] lb = 62.4 * (25 [ft]) A wc=12% 523 pounds per square feet. [pause] ft3 γT=117.7 [lb/ft3] lb = 1,560 ft2 lb σh=2,083 @ A lb σ’h = 523 ft2 ft2

? A Find: Ko σ’h σ’v γT=117.7 [lb/ft3] σh=2,083 Water Sand Ko= @ A lb 523 Ko= ft2 σ’v Water ? 10 [ft] Sand 15 [ft] 20 [ft] A wc=12% Next we solve for the vertical effective stress, at point A, γT=117.7 [lb/ft3] lb σh=2,083 @ A ft2

A Find: Ko σ’h σ’v σ’v = σv - u γT=117.7 [lb/ft3] σh=2,083 Water Sand 523 Ko= ft2 σ’v Water 10 [ft] σ’v = σv - u Sand 15 [ft] 20 [ft] A wc=12% which is equal to the total vertical stress, minus the pore water pressure. γT=117.7 [lb/ft3] lb σh=2,083 @ A ft2

A Find: Ko σ’h σ’v σ’v = σv - u γT=117.7 [lb/ft3] σh=2,083 Water Sand 523 Ko= ft2 σ’v Water 10 [ft] σ’v = σv - u lb 1,560 Sand 15 [ft] ft2 20 [ft] A wc=12% From before, we know the pore water pressure at point A is 1,560 pounds per square feet. γT=117.7 [lb/ft3] lb σh=2,083 @ A ft2

A Find: Ko σ’h σ’v σ’v = σv - u σv = Σ γ * d γT=117.7 [lb/ft3] 523 Ko= ft2 σ’v Water 10 [ft] σ’v = σv - u lb 1,560 Sand 15 [ft] ft2 σv = Σ γ * d 20 [ft] A wc=12% We next compute the total vertical stress by summing up the total unit weights times the depths, for each layer above point A, γT=117.7 [lb/ft3] lb σh=2,083 @ A ft2

A Find: Ko σ’h σ’v σ’v = σv - u σv = Σ γ * d σv = 62.4 lb 523 Ko= ft2 σ’v Water 10 [ft] σ’v = σv - u lb 1,560 Sand 15 [ft] ft2 σv = Σ γ * d 20 [ft] A lb wc=12% which is 62.4 pounds per cubic feet for water, times the 10 feet of water depth, σv = 62.4 * 10 [ft] γT=117.7 [lb/ft3] ft3 + … lb σh=2,083 @ A ft2

A Find: Ko σ’h σ’v σ’v = σv - u σv = Σ γ * d σv = 62.4 lb 523 Ko= ft2 σ’v Water 10 [ft] σ’v = σv - u lb 1,560 Sand 15 [ft] ft2 σv = Σ γ * d 20 [ft] A lb wc=12% plus, 117.7 pounds per cubic feet of the sand, times the 15 feet of sand depth. σv = 62.4 * 10 [ft] γT=117.7 [lb/ft3] ft3 lb + 117.7 * 15 [ft] lb σh=2,083 ft3 @ A ft2

A Find: Ko σ’h σ’v σ’v = σv - u σv = Σ γ * d σv = 62.4 lb 523 Ko= ft2 σ’v Water 10 [ft] σ’v = σv - u lb 1,560 Sand 15 [ft] ft2 σv = Σ γ * d 20 [ft] A lb wc=12% The total vertical stress is 2,390 pounds per square feet--- σv = 62.4 * 10 [ft] γT=117.7 [lb/ft3] ft3 lb + 117.7 * 15 [ft] lb σh=2,083 ft3 @ A lb ft2 σ’v = 2,390 ft2

A Find: Ko σ’h σ’v σ’v = σv - u σv = Σ γ * d σv = 62.4 lb 523 Ko= ft2 σ’v Water lb ft2 830 10 [ft] σ’v = σv - u lb 1,560 Sand 15 [ft] ft2 σv = Σ γ * d 20 [ft] A lb wc=12% which makes our total vertical effective stress equal to 830 pounds per square feet. σv = 62.4 * 10 [ft] γT=117.7 [lb/ft3] ft3 lb + 117.7 * 15 [ft] lb σh=2,083 ft3 @ A lb ft2 σ’v = 2,390 ft2

A Find: Ko σ’h σ’v γT=117.7 [lb/ft3] σh=2,083 Water Sand Ko= @ A lb 523 Ko= ft2 σ’v Water lb 10 [ft] 830 ft2 Sand 15 [ft] 20 [ft] A wc=12% Knowing the effective horizontal stress and the effective vertical stress, at point A, γT=117.7 [lb/ft3] lb σh=2,083 @ A ft2

A Find: Ko σ’h σ’v γT=117.7 [lb/ft3] σh=2,083 Water Sand Ko= Ko=0.63 523 Ko= ft2 σ’v Water lb 10 [ft] 830 ft2 Ko=0.63 Sand 15 [ft] 20 [ft] A wc=12% We compute the at-rest lateral earth pressure coefficient to be 0.63. γT=117.7 [lb/ft3] lb σh=2,083 @ A ft2

A Find: Ko σ’h σ’v γT=117.7 [lb/ft3] σh=2,083 Water Sand Ko= Ko=0.63 523 Ko= ft2 σ’v Water lb 10 [ft] 830 ft2 Ko=0.63 Sand 15 [ft] 20 [ft] 0.36 0.45 0.54 0.63 A wc=12% We revisit our possible solution, and determine the --- γT=117.7 [lb/ft3] lb σh=2,083 @ A ft2

A Find: Ko σ’h σ’v γT=117.7 [lb/ft3] σh=2,083 Water Sand Ko= Ko=0.63 523 Ko= ft2 σ’v Water lb 10 [ft] 830 ft2 Ko=0.63 Sand 15 [ft] AnswerD 20 [ft] 0.36 0.45 0.54 0.63 A wc=12% answer is D γT=117.7 [lb/ft3] lb σh=2,083 @ A ft2

? Index σ’v = Σ γ d γT=100 [lb/ft3] +γclay dclay 1 Find: σ’v at d = 30 feet (1+wc)*γw wc+(1/SG) σ’v = Σ γ d d Sand 10 ft γT=100 [lb/ft3] 100 [lb/ft3] 10 [ft] 20 ft Clay = γsand dsand +γclay dclay A W S V [ft3] W [lb] 40 ft text wc = 37% ? 20 [ft]