Find: ρc [in] from load γT=110 [lb/ft3] γT=100 [lb/ft3] Load=800 [lb/ft2] 1.5 3.0 C) 4.5 D) 6.0 CR=0.03 e=1.00 Clay CC=0.54 OCR=3.0 8 [ft] γT=110 [lb/ft3] Sand 4 [ft] γT=100 [lb/ft3] Find the settlement from consolidation in inches from a load. In this problem, Clay CR=0.04 e=0.80 CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3]

Find: ρc [in] from load γT=110 [lb/ft3] γT=100 [lb/ft3] Load=800 [lb/ft2] 1.5 3.0 C) 4.5 D) 6.0 CR=0.03 e=1.00 Clay CC=0.54 OCR=3.0 8 [ft] γT=110 [lb/ft3] Sand 4 [ft] γT=100 [lb/ft3] a vertical load of 800 pounds per square foot is applied to a soil profile consisting of --- Clay CR=0.04 e=0.80 CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3]

Find: ρc [in] from load γT=110 [lb/ft3] γT=100 [lb/ft3] Load=800 [lb/ft2] 1.5 3.0 C) 4.5 D) 6.0 e=1.00 Clay CR=0.03 CC=0.54 OCR=3.0 8 [ft] γT=110 [lb/ft3] Sand 4 [ft] γT=100 [lb/ft3] an 8 foot layer of clay, which is above --- Clay CR=0.04 e=0.80 CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3]

Find: ρc [in] from load γT=110 [lb/ft3] γT=100 [lb/ft3] Load=800 [lb/ft2] 1.5 3.0 C) 4.5 D) 6.0 e=1.00 Clay CR=0.03 CC=0.54 OCR=3.0 8 [ft] γT=110 [lb/ft3] Sand 4 [ft] γT=100 [lb/ft3] a four foot layer of sand, which is above --- Clay CR=0.04 e=0.80 CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3]

Find: ρc [in] from load γT=110 [lb/ft3] γT=100 [lb/ft3] Load=800 [lb/ft2] 1.5 3.0 C) 4.5 D) 6.0 e=1.00 Clay CR=0.03 CC=0.54 OCR=3.0 8 [ft] γT=110 [lb/ft3] Sand 4 [ft] γT=100 [lb/ft3] a 10 foot layer of clay. We’ve also been provided --- Clay CR=0.04 e=0.80 CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3]

Find: ρc [in] from load γT=110 [lb/ft3] γT=100 [lb/ft3] Load=800 [lb/ft2] 1.5 3.0 C) 4.5 D) 6.0 CR=0.03 e=1.00 Clay CC=0.54 OCR=3.0 8 [ft] γT=110 [lb/ft3] Sand γT=100 [lb/ft3] 4 [ft] various soil properties. In this problem, Clay CR=0.04 e=0.80 CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3]

Find: ρc [in] from load γT=110 [lb/ft3] γT=100 [lb/ft3] Load=800 [lb/ft2] CR=0.03 e=1.00 Clay CC=0.54 OCR=3.0 8 [ft] γT=110 [lb/ft3] Sand γT=100 [lb/ft3] 4 [ft] only the clay layers will settle from consolidation. For both clay layers, we’ve been provided the --- Clay CR=0.04 e=0.80 CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3]

Find: ρc [in] from load γT=110 [lb/ft3] γT=100 [lb/ft3] Load=800 [lb/ft2] CR=0.03 e=1.00 Clay CC=0.54 OCR=3.0 8 [ft] γT=110 [lb/ft3] Sand γT=100 [lb/ft3] 4 [ft] the over consolidation ratio, Clay CR=0.04 e=0.80 CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3]

Find: ρc [in] from load γT=110 [lb/ft3] γT=100 [lb/ft3] Load=800 [lb/ft2] CR=0.03 e=1.00 Clay CC=0.54 OCR=3.0 8 [ft] γT=110 [lb/ft3] Sand γT=100 [lb/ft3] 4 [ft] the recompression index, Clay CR=0.04 e=0.80 CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3]

Find: ρc [in] from load γT=110 [lb/ft3] γT=100 [lb/ft3] Load=800 [lb/ft2] CR=0.03 e=1.00 Clay CC=0.54 OCR=3.0 8 [ft] γT=110 [lb/ft3] Sand γT=100 [lb/ft3] 4 [ft] and the compression index, sometimes called the virgin compression index. Soil settlement can be plotted --- Clay CR=0.04 e=0.80 CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3]

Find: ρc [in] from load eo γT=110 [lb/ft3] γT=100 [lb/ft3] Load=800 [lb/ft2] e=1.00 log σ’v eo Clay CR=0.03 CC=0.54 OCR=3.0 8 [ft] γT=110 [lb/ft3] Sand γT=100 [lb/ft3] 4 [ft] on a graph where the vertical axis --- Clay CR=0.04 e=0.80 CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3]

Find: ρc [in] from load eo γT=110 [lb/ft3] γT=100 [lb/ft3] Load=800 [lb/ft2] e=1.00 log σ’v eo Clay CR=0.03 CC=0.54 OCR=3.0 8 [ft] γT=110 [lb/ft3] Sand γT=100 [lb/ft3] 4 [ft] is the void ratio, and the horizontal axis is --- Clay CR=0.04 e=0.80 CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3]

Find: ρc [in] from load eo γT=110 [lb/ft3] γT=100 [lb/ft3] Load=800 [lb/ft2] e=1.00 log σ’v eo Clay CR=0.03 CC=0.54 OCR=3.0 8 [ft] γT=110 [lb/ft3] Sand γT=100 [lb/ft3] 4 [ft] Log of the vertical effective stress. The preconsolidated stress of the soil occurs --- Clay CR=0.04 e=0.80 CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3]

Find: ρc [in] from load eo σ’p γT=110 [lb/ft3] γT=100 [lb/ft3] Load=800 [lb/ft2] e=1.00 log σ’v eo Clay CR=0.03 CC=0.54 OCR=3.0 8 [ft] γT=110 [lb/ft3] Sand γT=100 [lb/ft3] 4 [ft] where the curve changes slope. There is an graphical technique used to determine the preconsolidation stress, Clay CR=0.04 e=0.80 σ’p CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3]

Find: ρc [in] from load eo σ’p γT=110 [lb/ft3] γT=100 [lb/ft3] Load=800 [lb/ft2] e=1.00 log σ’v eo Clay CR=0.03 CC=0.54 OCR=3.0 8 [ft] γT=110 [lb/ft3] Sand γT=100 [lb/ft3] 4 [ft] but I’ll skip that for now. If the soil is normally consolidated, the point on the curve would be here, Clay CR=0.04 e=0.80 σ’p CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3]

Find: ρc [in] from load eo σ’p γT=110 [lb/ft3] γT=100 [lb/ft3] Load=800 [lb/ft2] recompression e=1.00 log σ’v eo Clay CR=0.03 CC=0.54 OCR=3.0 8 [ft] γT=110 [lb/ft3] Sand γT=100 [lb/ft3] 4 [ft] with the recompression curve to the left of the point, Clay CR=0.04 e=0.80 σ’p CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3]

Find: ρc [in] from load eo σ’p γT=110 [lb/ft3] γT=100 [lb/ft3] Load=800 [lb/ft2] recompression virgin compression e=1.00 log σ’v eo Clay CR=0.03 CC=0.54 OCR=3.0 8 [ft] γT=110 [lb/ft3] Sand γT=100 [lb/ft3] 4 [ft] and the virgin compression curve to the right. Since we notice that ---- Clay CR=0.04 e=0.80 σ’p CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3]

Find: ρc [in] from load eo σ’p γT=110 [lb/ft3] γT=100 [lb/ft3] Load=800 [lb/ft2] recompression virgin compression e=1.00 log σ’v eo Clay CR=0.03 CC=0.54 OCR=3.0 8 [ft] γT=110 [lb/ft3] Sand γT=100 [lb/ft3] 4 [ft] both layers of clay are over consolidated, they would both begin --- Clay CR=0.04 e=0.80 σ’p CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3]

Find: ρc [in] from load eo σ’p γT=110 [lb/ft3] γT=100 [lb/ft3] Load=800 [lb/ft2] recompression e=1.00 log σ’v eo Clay CR=0.03 CC=0.54 OCR=3.0 8 [ft] γT=110 [lb/ft3] Sand γT=100 [lb/ft3] 4 [ft] on the recompression curve. [pause] The final vertical effective stress on the soil --- Clay CR=0.04 e=0.80 σ’p CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3]

Find: ρc [in] from load eo σ’i σ’p σ’final = σ’initial +Load Load=800 [lb/ft2] e=1.00 log σ’v eo Clay CR=0.03 CC=0.54 OCR=3.0 8 [ft] γT=110 [lb/ft3] Sand γT=100 [lb/ft3] 4 [ft] equals the initial vertical effective stress --- Clay CR=0.04 e=0.80 σ’i σ’p CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3] σ’final = σ’initial +Load

Find: ρc [in] from load eo σ’i σ’p σ’final = σ’initial +Load Load=800 [lb/ft2] e=1.00 log σ’v eo Clay CR=0.03 CC=0.54 OCR=3.0 8 [ft] γT=110 [lb/ft3] Sand γT=100 [lb/ft3] 4 [ft] plus the load. Clay CR=0.04 e=0.80 σ’i σ’p CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3] σ’final = σ’initial +Load

Find: ρc [in] from load eo σ’i σ’p σ’final = σ’initial +Load Load=800 [lb/ft2] e=1.00 log σ’v eo Clay CR=0.03 CC=0.54 OCR=3.0 8 [ft] γT=110 [lb/ft3] Sand γT=100 [lb/ft3] 4 [ft] If the final stress is less than --- Clay CR=0.04 e=0.80 σ’i σ’p CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3] σ’final = σ’initial +Load

Find: ρc [in] from load eo σ’i σ’f σ’p γT=110 [lb/ft3] γT=100 [lb/ft3] Load=800 [lb/ft2] log σ’v eo Clay CR=0.03 e=1.00 CC=0.54 OCR=3.0 8 [ft] γT=110 [lb/ft3] Sand γT=100 [lb/ft3] 4 [ft] the preconsolidation stress, Clay CR=0.04 e=0.80 σ’i σ’f σ’p CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3]

Find: ρc [in] from load eo σ’i σ’f σ’p σ’i < σ’f < σ’p Load=800 [lb/ft2] CR=0.03 e=1.00 log σ’v eo Clay CC=0.54 OCR=3.0 8 [ft] γT=110 [lb/ft3] Sand γT=100 [lb/ft3] 4 [ft] then the entire consolidation occurs along the recompression curve, Clay CR=0.04 e=0.80 σ’i σ’f σ’p CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3] σ’i < σ’f < σ’p

Find: ρc [in] from load eo σ’i σ’f σ’p σ’i < σ’f < σ’p Load=800 [lb/ft2] CR=0.03 e=1.00 log σ’v eo Clay CC=0.54 OCR=3.0 8 [ft] γT=110 [lb/ft3] Sand γT=100 [lb/ft3] 4 [ft] which has a slope equal in magnitude to the recompression index, Clay CR=0.04 e=0.80 σ’i σ’f σ’p CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3] σ’i < σ’f < σ’p

( ) Find: ρc [in] from load H*C σfinal ρc= log 1+e σinitial Load=800 [lb/ft2] ( ) H*C σfinal ρc= 1+e σinitial log ‘ e=1.00 Clay CR=0.03 CC=0.54 OCR=3.0 8 [ft] γT=110 [lb/ft3] Sand γT=100 [lb/ft3] 4 [ft] If this is the case, we would use the recompression index --- Clay CR=0.04 e=0.80 CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3] σ’i < σ’f < σ’p

( ) ( ) Find: ρc [in] from load H*C σfinal ρc= log 1+e σinitial Load=800 [lb/ft2] ( ) H*C σfinal ρc= 1+e σinitial log ‘ Clay CR=0.03 e=1.00 CC=0.54 OCR=3.0 8 [ft] γT=110 [lb/ft3] H*CR σfinal ( ) ‘ ρc= log Sand 1+e σinitial γT=100 [lb/ft3] in the consolidation settlement equation. [pause] However, --- ‘ 4 [ft] Clay CR=0.04 e=0.80 CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3] σ’i < σ’f < σ’p

Find: ρc [in] from load eo σ’i σ’p γT=110 [lb/ft3] γT=100 [lb/ft3] Load=800 [lb/ft2] e=1.00 log σ’v eo Clay CR=0.03 CC=0.54 OCR=3.0 8 [ft] γT=110 [lb/ft3] Sand γT=100 [lb/ft3] 4 [ft] if the final stress exceeds --- Clay CR=0.04 e=0.80 σ’i σ’p CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3]

Find: ρc [in] from load eo σ’i σ’p σ’f γT=110 [lb/ft3] γT=100 [lb/ft3] Load=800 [lb/ft2] log σ’v eo Clay CR=0.03 e=1.00 CC=0.54 OCR=3.0 8 [ft] γT=110 [lb/ft3] Sand γT=100 [lb/ft3] 4 [ft] the preconsolidation stress, Clay CR=0.04 e=0.80 σ’i σ’p σ’f CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3]

Find: ρc [in] from load eo σ’i σ’p σ’f σ’i < σ’p < σ’f Load=800 [lb/ft2] CR=0.03 e=1.00 log σ’v eo Clay CC=0.54 OCR=3.0 8 [ft] γT=110 [lb/ft3] Sand γT=100 [lb/ft3] 4 [ft] we’ll end up evaluating --- Clay CR=0.04 e=0.80 σ’i σ’p σ’f CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3] σ’i < σ’p < σ’f

Find: ρc [in] from load eo σ’i σ’p σ’f σ’i < σ’p < σ’f Load=800 [lb/ft2] CR=0.03 e=1.00 log σ’v eo Clay CC=0.54 OCR=3.0 8 [ft] γT=110 [lb/ft3] Sand γT=100 [lb/ft3] 4 [ft] part of the settlement along the recompression curve, and part of the settlement --- Clay CR=0.04 e=0.80 σ’i σ’p σ’f CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3] σ’i < σ’p < σ’f

Find: ρc [in] from load eo σ’i σ’p σ’f σ’i < σ’p < σ’f Load=800 [lb/ft2] CR=0.03 e=1.00 log σ’v eo Clay CC=0.54 OCR=3.0 8 [ft] γT=110 [lb/ft3] Sand γT=100 [lb/ft3] 4 [ft] along the virgin compression curve. When solving for this second settlement, --- Clay CR=0.04 e=0.80 σ’i σ’p σ’f CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3] σ’i < σ’p < σ’f

( ) ( ) ( ) + Find: ρc [in] from load H*C σfinal ρc= log 1+e σinitial Load=800 [lb/ft2] ( ) H*C σfinal ρc= 1+e σinitial log ‘ Clay CR=0.03 e=1.00 CC=0.54 OCR=3.0 8 [ft] γT=110 [lb/ft3] H*CR σp ( ) ‘ ρc= log Sand 1+e σinitial ‘ γT=100 [lb/ft3] 4 [ft] we compute the two consolidation settlements separately, then add them together. Let’s begin with the 8 foot layer of clay. H*CC σfinal ( ) ‘ + Clay CR=0.04 e=0.80 log 1+e σp ‘ CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3]

Find: ρc [in] from load σ’i = γ’clay* dclay γT=110 [lb/ft3] Load=800 [lb/ft2] σ’i = γ’clay* dclay e=1.00 Clay CR=0.03 CC=0.54 OCR=3.0 8 [ft] γT=110 [lb/ft3] Sand γT=100 [lb/ft3] 4 [ft] The initial effective stress is equal to the effective unit weight times the depth. Clay CR=0.04 e=0.80 CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3]

Find: ρc [in] from load σ’i = γ’clay* dclay γT=110 [lb/ft3] Load=800 [lb/ft2] 110 [lb/ft3] σ’i = γ’clay* dclay e=1.00 Clay CR=0.03 CC=0.54 OCR=3.0 8 [ft] γT=110 [lb/ft3] Sand γT=100 [lb/ft3] 4 [ft] Since the groundwater table is deep, the effective unit weight equals 110 pounds per cubic foot. Clay CR=0.04 e=0.80 CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3]

Find: ρc [in] from load σ’i = γ’clay* dclay γT=110 [lb/ft3] Load=800 [lb/ft2] 110 [lb/ft3] 4 [ft] σ’i = γ’clay* dclay CR=0.03 e=1.00 Clay CC=0.54 OCR=3.0 8 [ft] γT=110 [lb/ft3] Sand γT=100 [lb/ft3] 4 [ft] If we don’t subdivide the layer, we’ll set the depth equal to the middle of the layer, so, ‘d’ equals 4 feet. Clay CR=0.04 e=0.80 CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3]

Find: ρc [in] from load σ’i = γ’clay* dclay γT=110 [lb/ft3] Load=800 [lb/ft2] 110 [lb/ft3] 4 [ft] σ’i = γ’clay* dclay e=1.00 Clay CR=0.03 CC=0.54 OCR=3.0 = 440 [lb/ft2] 8 [ft] γT=110 [lb/ft3] Sand γT=100 [lb/ft3] 4 [ft] our initial stress is 440 pounds per square feet. Clay CR=0.04 e=0.80 CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3]

Find: ρc [in] from load σ’i = γ’clay* dclay σ’f = σ’i + load Load=800 [lb/ft2] 110 [lb/ft3] 4 [ft] σ’i = γ’clay* dclay e=1.00 Clay CR=0.03 CC=0.54 OCR=3.0 = 440 [lb/ft2] 8 [ft] γT=110 [lb/ft3] σ’f = σ’i + load Sand γT=100 [lb/ft3] 4 [ft] And after adding the load, --- Clay CR=0.04 e=0.80 CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3]

Find: ρc [in] from load σ’i = γ’clay* dclay σ’f = σ’i + load Load=800 [lb/ft2] 110 [lb/ft3] 4 [ft] σ’i = γ’clay* dclay e=1.00 Clay CR=0.03 CC=0.54 OCR=3.0 = 440 [lb/ft2] 8 [ft] γT=110 [lb/ft3] σ’f = σ’i + load Sand γT=100 [lb/ft3] 4 [ft] our final stress equals --- Clay CR=0.04 e=0.80 CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3]

Find: ρc [in] from load σ’i = γ’clay* dclay σ’f = σ’i + load Load=800 [lb/ft2] 110 [lb/ft3] 4 [ft] σ’i = γ’clay* dclay e=1.00 Clay CR=0.03 CC=0.54 OCR=3.0 = 440 [lb/ft2] 8 [ft] γT=110 [lb/ft3] σ’f = σ’i + load Sand γT=100 [lb/ft3] 4 [ft] 1,240 pounds per square feet. [pause] = 1,240 [lb/ft2] Clay CR=0.04 e=0.80 CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3]

Find: ρc [in] from load σ’i = γ’clay* dclay σ’f = σ’i + load Load=800 [lb/ft2] 110 [lb/ft3] 4 [ft] σ’i = γ’clay* dclay e=1.00 Clay CR=0.03 CC=0.54 OCR=3.0 = 440 [lb/ft2] 8 [ft] γT=110 [lb/ft3] σ’f = σ’i + load Sand γT=100 [lb/ft3] 4 [ft] And our preconsolidation stress is --- = 1,240 [lb/ft2] σ’p = σ’i * OCR Clay CR=0.04 e=0.80 CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3]

Find: ρc [in] from load σ’i = γ’clay* dclay σ’f = σ’i + load Load=800 [lb/ft2] 110 [lb/ft3] 4 [ft] σ’i = γ’clay* dclay e=1.00 Clay CR=0.03 CC=0.54 OCR=3.0 = 440 [lb/ft2] 8 [ft] γT=110 [lb/ft3] σ’f = σ’i + load Sand γT=100 [lb/ft3] 4 [ft] the initial stress times the overconsolidation ratio, or, = 1,240 [lb/ft2] σ’p = σ’i * OCR Clay CR=0.04 e=0.80 CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3]

Find: ρc [in] from load σ’i = γ’clay* dclay σ’f = σ’i + load Load=800 [lb/ft2] 110 [lb/ft3] 4 [ft] σ’i = γ’clay* dclay e=1.00 Clay CR=0.03 CC=0.54 OCR=3.0 = 440 [lb/ft2] 8 [ft] γT=110 [lb/ft3] σ’f = σ’i + load Sand γT=100 [lb/ft3] 4 [ft] 1,320 pounds per square feet. [pause] Since the final stress --- = 1,240 [lb/ft2] σ’p = σ’i * OCR Clay CR=0.04 e=0.80 = 1,320 [lb/ft2] CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3]

Find: ρc [in] from load σ’i = γ’clay* dclay σ’f = σ’i + load Load=800 [lb/ft2] 110 [lb/ft3] 4 [ft] σ’i = γ’clay* dclay e=1.00 Clay CR=0.03 CC=0.54 OCR=3.0 = 440 [lb/ft2] 8 [ft] γT=110 [lb/ft3] σ’f = σ’i + load Sand γT=100 [lb/ft3] 4 [ft] is less than the preconsolidated stress,--- = 1,240 [lb/ft2] σ’p = σ’i * OCR Clay CR=0.04 e=0.80 = 1,320 [lb/ft2] CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3] σ’i < σ’f < σ’p

Find: ρc [in] from load σ’i = 440 [lb/ft2] σ’f = 1,240 [lb/ft2] Load=800 [lb/ft2] σ’i = 440 [lb/ft2] e=1.00 Clay CR=0.03 CC=0.54 OCR=3.0 8 [ft] γT=110 [lb/ft3] σ’f = 1,240 [lb/ft2] Sand γT=100 [lb/ft3] 4 [ft] the settlement at a depth of 4 feet is only along the recompression curve. σ’p = 1,320 [lb/ft2] Clay CR=0.04 e=0.80 CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3] σ’i < σ’f < σ’p

Find: ρc [in] from load σ’i = 440 [lb/ft2] σ’f = 1,240 [lb/ft2] Load=800 [lb/ft2] CR=0.03 e=1.00 Clay CC=0.54 OCR=3.0 8 [ft] γT=110 [lb/ft3] Sand γT=100 [lb/ft3] 4 [ft] For this problem, we’ll assume this is true for the entire 8 foot thick layer of clay. σ’i = 440 [lb/ft2] Clay CR=0.04 e=0.80 σ’f = 1,240 [lb/ft2] CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3] σ’i < σ’f < σ’p

( ) Find: ρc [in] from load H*CR σfinal ρc= log 1+e σinitial Load=800 [lb/ft2] e=1.00 Clay CR=0.03 H*CR σfinal ( ) ‘ ρc= log CC=0.54 OCR=3.0 1+e σinitial 8 [ft] ‘ γT=110 [lb/ft3] Sand γT=100 [lb/ft3] 4 [ft] So we substitute in our known values --- σ’i = 440 [lb/ft2] Clay CR=0.04 e=0.80 σ’f = 1,240 [lb/ft2] CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3] σ’i < σ’f < σ’p

( ) Find: ρc [in] from load H*CR σfinal ρc= log 1+e σinitial Load=800 [lb/ft2] e=1.00 Clay CR=0.03 H*CR σfinal ( ) ‘ ρc= log CC=0.54 OCR=3.0 1+e σinitial 8 [ft] ‘ γT=110 [lb/ft3] Sand γT=100 [lb/ft3] 4 [ft] and find the settlement in the top clay layer is σ’i = 440 [lb/ft2] Clay CR=0.04 e=0.80 σ’f = 1,240 [lb/ft2] CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3] σ’i < σ’f < σ’p

( ) Find: ρc [in] from load H*CR σfinal ρc= log 1+e σinitial Load=800 [lb/ft2] e=1.00 Clay CR=0.03 H*CR σfinal ( ) ‘ ρc= log CC=0.54 OCR=3.0 1+e σinitial 8 [ft] ‘ γT=110 [lb/ft3] Sand ρc=0.65[in] γT=100 [lb/ft3] 4 [ft] 0.65 inches. [pause] σ’i = 440 [lb/ft2] Clay CR=0.04 e=0.80 σ’f = 1,240 [lb/ft2] CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3] σ’i < σ’f < σ’p

Find: ρc [in] from load γT=110 [lb/ft3] γT=100 [lb/ft3] Load=800 [lb/ft2] CR=0.03 e=1.00 Clay CC=0.54 OCR=3.0 8 [ft] γT=110 [lb/ft3] Sand γT=100 [lb/ft3] 4 [ft] Next we’ll evaluate the settlement in the 10 foot thick clay layer. Clay CR=0.04 e=0.80 CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3]

Find: ρc [in] from load σ’i = Σ γ’ * d γT=110 [lb/ft3] γT=100 [lb/ft3] Load=800 [lb/ft2] σ’i = Σ γ’ * d e=1.00 Clay CR=0.03 CC=0.54 OCR=3.0 8 [ft] γT=110 [lb/ft3] Sand γT=100 [lb/ft3] 4 [ft] As before, Clay CR=0.04 e=0.80 CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3]

Find: ρc [in] from load σ’i = Σ γ’ * d σ’i = γ’clay,top*dclay,top Load=800 [lb/ft2] σ’i = Σ γ’ * d e=1.00 σ’i = γ’clay,top*dclay,top Clay CR=0.03 CC=0.54 OCR=3.0 + γ’sand*dsand 8 [ft] γT=110 [lb/ft3] + γ’clay,bot*dclay,bot Sand γT=100 [lb/ft3] 4 [ft] we determine the vertical effective stress the the midpoint of the clay layer Clay CR=0.04 e=0.80 CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3]

Find: ρc [in] from load σ’i = Σ γ’ * d σ’i = γ’clay,top*dclay,top Load=800 [lb/ft2] σ’i = Σ γ’ * d e=1.00 σ’i = γ’clay,top*dclay,top Clay CR=0.03 CC=0.54 OCR=3.0 + γ’sand*dsand 8 [ft] γT=110 [lb/ft3] + γ’clay,bot*dclay,bot Sand γT=100 [lb/ft3] 4 [ft] Where the depth into the 10 foot clay layer will be calculated with ‘d’ equal to 5 feet. 5[ft] Clay CR=0.04 e=0.80 CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3]

Find: ρc [in] from load σ’i = Σ γ’ * d σ’i = γ’clay,top*dclay,top Load=800 [lb/ft2] σ’i = Σ γ’ * d e=1.00 σ’i = γ’clay,top*dclay,top Clay CR=0.03 CC=0.54 OCR=3.0 + γ’sand*dsand 8 [ft] γT=110 [lb/ft3] + γ’clay,bot*dclay,bot Sand γT=100 [lb/ft3] 4 [ft] The initial stress is 1,880 pounds per square foot. 5[ft] σ’i = 1,880[lb/ft2] Clay CR=0.04 e=0.80 CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3]

Find: ρc [in] from load σ’i = 1,880[lb/ft2] γT=110 [lb/ft3] Load=800 [lb/ft2] CR=0.03 e=1.00 Clay CC=0.54 OCR=3.0 8 [ft] γT=110 [lb/ft3] Sand γT=100 [lb/ft3] 4 [ft] --- σ’i = 1,880[lb/ft2] Clay CR=0.04 e=0.80 CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3]

Find: ρc [in] from load σ’i = 1,880[lb/ft2] σ’f = σ’i + load Load=800 [lb/ft2] σ’i = 1,880[lb/ft2] e=1.00 Clay CR=0.03 CC=0.54 OCR=3.0 8 [ft] σ’f = σ’i + load γT=110 [lb/ft3] Sand γT=100 [lb/ft3] 4 [ft] After adding the load, --- Clay CR=0.04 e=0.80 CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3]

Find: ρc [in] from load σ’i = 1,880[lb/ft2] σ’f = σ’i + load Load=800 [lb/ft2] σ’i = 1,880[lb/ft2] e=1.00 Clay CR=0.03 CC=0.54 OCR=3.0 8 [ft] σ’f = σ’i + load γT=110 [lb/ft3] Sand γT=100 [lb/ft3] 4 [ft] we calculate the final stress value to be --- Clay CR=0.04 e=0.80 CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3]

Find: ρc [in] from load σ’i = 1,880[lb/ft2] σ’f = σ’i + load Load=800 [lb/ft2] σ’i = 1,880[lb/ft2] e=1.00 Clay CR=0.03 CC=0.54 OCR=3.0 8 [ft] σ’f = σ’i + load γT=110 [lb/ft3] = 2,680[lb/ft2] Sand γT=100 [lb/ft3] 4 [ft] 2,680 pounds per square feet. [pause] For this lower clay layer, --- Clay CR=0.04 e=0.80 CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3]

Find: ρc [in] from load σ’i = 1,880[lb/ft2] σ’f = σ’i + load Load=800 [lb/ft2] σ’i = 1,880[lb/ft2] e=1.00 Clay CR=0.03 CC=0.54 OCR=3.0 8 [ft] σ’f = σ’i + load γT=110 [lb/ft3] = 2,680[lb/ft2] Sand γT=100 [lb/ft3] 4 [ft] the pre-consolitation stress calculates to ---- σ’p = σ’i * OCR Clay CR=0.04 e=0.80 CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3]

Find: ρc [in] from load σ’i = 1,880[lb/ft2] σ’f = σ’i + load Load=800 [lb/ft2] σ’i = 1,880[lb/ft2] e=1.00 Clay CR=0.03 CC=0.54 OCR=3.0 8 [ft] σ’f = σ’i + load γT=110 [lb/ft3] = 2,680[lb/ft2] Sand γT=100 [lb/ft3] 4 [ft] 2,256 pounds per square feet. This means, the final stress exceeds --- σ’p = σ’i * OCR = 2,256[lb/ft2] Clay CR=0.04 e=0.80 CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3]

Find: ρc [in] from load σ’i = 1,880[lb/ft2] σ’f = σ’i + load Load=800 [lb/ft2] σ’i = 1,880[lb/ft2] e=1.00 Clay CR=0.03 CC=0.54 OCR=3.0 8 [ft] σ’f = σ’i + load γT=110 [lb/ft3] = 2,680[lb/ft2] Sand γT=100 [lb/ft3] 4 [ft] the preconsolidated stress of the soil, which means this 10 foot thick clay layer of clay --- σ’p = σ’i * OCR = 2,256[lb/ft2] Clay CR=0.04 e=0.80 CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3] σ’i < σ’p < σ’f

( ) ( ) Find: ρc [in] from load H*CR σp ρc= log 1+e σinitial Load=800 [lb/ft2] H*CR σp ( ) ‘ ρc= log 1+e σinitial CR=0.03 e=1.00 Clay CC=0.54 OCR=3.0 H*CC σfinal ( ) ‘ 8 [ft] γT=110 [lb/ft3] log 1+e σp ‘ Sand γT=100 [lb/ft3] σ’i = 1,880[lb/ft2] 4 [ft] will consolidate along the recompression curve AND the virgin compression curve. After substituting in, --- σ’f = 2,680[lb/ft2] Clay CR=0.04 e=0.80 σ’p = 2,256[lb/ft2] CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3] σ’i < σ’p < σ’f

( ) ( ) Find: ρc [in] from load H*CR σp ρc= log 1+e σinitial Load=800 [lb/ft2] H*CR σp ( ) ‘ ρc= log 1+e σinitial e=1.00 Clay CR=0.03 CC=0.54 OCR=3.0 H*CC σfinal ( ) ‘ 8 [ft] γT=110 [lb/ft3] log 1+e σp ‘ Sand γT=100 [lb/ft3] σ’i = 1,880[lb/ft2] 4 [ft] all the required values, we calculate the consolidation settlement equals --- σ’f = 2,680[lb/ft2] Clay CR=0.04 e=0.80 σ’p = 2,256[lb/ft2] CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3] σ’i < σ’p < σ’f

Find: ρc [in] from load ρc= 0.21[in]+3.59[in] σ’i = 1,880[lb/ft2] Load=800 [lb/ft2] ‘ ρc= 0.21[in]+3.59[in] e=1.00 Clay CR=0.03 CC=0.54 OCR=3.0 8 [ft] γT=110 [lb/ft3] Sand γT=100 [lb/ft3] σ’i = 1,880[lb/ft2] 4 [ft] 0.21 inches along the recompression curve and 3.59 inches along the virgin compression curve. For a total of ---- σ’f = 2,680[lb/ft2] Clay CR=0.04 e=0.80 σ’p = 2,256[lb/ft2] CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3] σ’i < σ’p < σ’f

Find: ρc [in] from load ρc= 0.21[in]+3.59[in] σ’i = 1,880[lb/ft2] Load=800 [lb/ft2] ‘ ρc= 0.21[in]+3.59[in] CR=0.03 e=1.00 Clay = 3.80[in] CC=0.54 OCR=3.0 8 [ft] γT=110 [lb/ft3] Sand γT=100 [lb/ft3] σ’i = 1,880[lb/ft2] 4 [ft] 3.80 inches of consolidation settlement. [pause] σ’f = 2,680[lb/ft2] Clay CR=0.04 e=0.80 σ’p = 2,256[lb/ft2] CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3] σ’i < σ’p < σ’f

Find: ρc [in] from load ρc,total =ρc,top+ρc,bottom σ’i = 1,880[lb/ft2] Load=800 [lb/ft2] ‘ e=1.00 Clay CR=0.03 ρc,total =ρc,top+ρc,bottom CC=0.54 OCR=3.0 8 [ft] γT=110 [lb/ft3] Sand γT=100 [lb/ft3] σ’i = 1,880[lb/ft2] 4 [ft] After adding the settlement --- σ’f = 2,680[lb/ft2] Clay CR=0.04 e=0.80 σ’p = 2,256[lb/ft2] CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3] σ’i < σ’p < σ’f

Find: ρc [in] from load ρc,total =ρc,top+ρc,bottom σ’i = 1,880[lb/ft2] Load=800 [lb/ft2] ‘ 3.80[in] 0.65[in] CR=0.03 e=1.00 Clay ρc,total =ρc,top+ρc,bottom CC=0.54 OCR=3.0 8 [ft] γT=110 [lb/ft3] Sand γT=100 [lb/ft3] σ’i = 1,880[lb/ft2] 4 [ft] from both clay layers together, σ’f = 2,680[lb/ft2] Clay CR=0.04 e=0.80 σ’p = 2,256[lb/ft2] CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3] σ’i < σ’p < σ’f

Find: ρc [in] from load ρc,total =ρc,top+ρc,bottom = 4.45 [in] Load=800 [lb/ft2] ‘ 3.80[in] 0.65[in] e=1.00 Clay CR=0.03 ρc,total =ρc,top+ρc,bottom CC=0.54 OCR=3.0 = 4.45 [in] 8 [ft] γT=110 [lb/ft3] Sand γT=100 [lb/ft3] σ’i = 1,880[lb/ft2] 4 [ft] We find the total settlement is 4.45 inches. σ’f = 2,680[lb/ft2] Clay CR=0.04 e=0.80 σ’p = 2,256[lb/ft2] CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3] σ’i < σ’p < σ’f

Find: ρc [in] from load ρc,total =ρc,top+ρc,bottom = 4.45 [in] Load=800 [lb/ft2] ‘ 3.80[in] 0.65[in] e=1.00 Clay CR=0.03 ρc,total =ρc,top+ρc,bottom CC=0.54 OCR=3.0 = 4.45 [in] 8 [ft] γT=110 [lb/ft3] Sand 1.5 3.0 C) 4.5 D) 6.0 γT=100 [lb/ft3] σ’i = 1,880[lb/ft2] 4 [ft] So we return to our possible solutions σ’f = 2,680[lb/ft2] Clay CR=0.04 e=0.80 σ’p = 2,256[lb/ft2] CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3] σ’i < σ’p < σ’f

Find: ρc [in] from load ρc,total =ρc,top+ρc,bottom = 4.45 [in] Load=800 [lb/ft2] ‘ 3.80[in] 0.65[in] CR=0.03 e=1.00 Clay ρc,total =ρc,top+ρc,bottom CC=0.54 OCR=3.0 = 4.45 [in] 8 [ft] γT=110 [lb/ft3] Answer: C Sand 1.5 3.0 C) 4.5 D) 6.0 γT=100 [lb/ft3] σ’i = 1,880[lb/ft2] 4 [ft] and find the answer is C σ’f = 2,680[lb/ft2] Clay CR=0.04 e=0.80 σ’p = 2,256[lb/ft2] CC=0.72 OCR=1.2 10[ft] γT=120 [lb/ft3] σ’i < σ’p < σ’f

( ) ? γclay=53.1[lb/ft3] Index σ’v = Σ γ d H*C σfinal ρcn= log Find: σ’v ρc 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% ? Δh 20 [ft] (5 [cm])2 * π/4 ( ) H*C σfinal ρcn= 1+e σinitial log ‘