Find: ρc [in] from load (4 layers)

Find: ρc [in] from load (4 layers)
Load=200[lb/ft2] 1.0 2.5 C) 5 D) 7.5 Sand 10 ft γT=100 [lb/ft3] Clay 40 ft Find the consolidation settlement of the clay, using 4 layers. In this problem, wc = 37% Cc=0.72 OCR=1.0

Load=200[lb/ft2] 1.0 2.5 C) 5 D) 7.5 Sand 10 ft γT=100 [lb/ft3] Clay 40 ft an incremental load of 200 pounds per square foot is added to a soil profile -- wc = 37% Cc=0.72 OCR=1.0

Load=200[lb/ft2] 1.0 2.5 C) 5 D) 7.5 Sand 10 ft γT=100 [lb/ft3] Clay 40 ft consisting of a 10 foot layer of sand, wc = 37% Cc=0.72 OCR=1.0

Load=200[lb/ft2] 1.0 2.5 C) 5 D) 7.5 Sand 10 ft γT=100 [lb/ft3] Clay 40 ft and a 40 foot layer of clay. wc = 37% Cc=0.72 OCR=1.0

Load=200[lb/ft2] 1.0 2.5 C) 5 D) 7.5 Sand 10 ft γT=100 [lb/ft3] Clay 40 ft We’ve also been given some soil properties. And we notice --- wc = 37% Cc=0.72 OCR=1.0

Load=200[lb/ft2] 1.0 2.5 C) 5 D) 7.5 Sand 10 ft γT=100 [lb/ft3] Clay 40 ft the groundwater table is at the sand-clay interface. wc = 37% Cc=0.72 OCR=1.0

Load=200[lb/ft2] 1.0 2.5 C) 5 D) 7.5 Sand 10 ft γT=100 [lb/ft3] Clay 40 ft --- wc = 37% Cc=0.72 OCR=1.0

Load=200[lb/ft2] Sand 10 ft γT=100 [lb/ft3] Clay 40 ft The generalized equation for --- wc = 37% Cc=0.72 OCR=1.0

( ) Find: ρc [in] from load (4 layers) H*C σ’final ρc= log
Load=200[lb/ft2] Sand 10 ft H*C σ’final ( ) ρc= 1+e σinitial log γT=100 [lb/ft3] ‘ ‘ Clay 40 ft consolidation settlement is wc = 37% Cc=0.72 OCR=1.0

( ) Find: ρc [in] from load (4 layers) H*C σfinal ρc= log 1+e σinitial
Load=200[lb/ft2] Sand 10 ft H*C σfinal ( ) ρc= 1+e σinitial log γT=100 [lb/ft3] ‘ ‘ Clay 40 ft the height, or thickness, of the layer, wc = 37% Cc=0.72 OCR=1.0

Load=200[lb/ft2] Sand 10 ft H*C σfinal ( ) ρc= 1+e σinitial log γT=100 [lb/ft3] ‘ ‘ Clay 40 ft times the compression index, wc = 37% Cc=0.72 OCR=1.0

Load=200[lb/ft2] Sand 10 ft H*C σfinal ( ) ρc= 1+e σinitial log γT=100 [lb/ft3] ‘ ‘ Clay 40 ft divided by 1 plus the initial void ratio, wc = 37% Cc=0.72 OCR=1.0

Load=200[lb/ft2] Sand 10 ft H*C σfinal ( ) ρc= 1+e σinitial log γT=100 [lb/ft3] ‘ ‘ Clay 40 ft all multiplied by the logarithm of the final vertical effective stress divided by the initial vertical effective stress. [pause] wc = 37% Cc=0.72 OCR=1.0

Load=200[lb/ft2] Sand 10 ft H*C σfinal ( ) ρc= 1+e σinitial log γT=100 [lb/ft3] ‘ ‘ Clay 40 ft Notice the over consolidation ratio in the clay layer is 1, so, the clay is normally consolidated, wc = 37% Cc=0.72 OCR=1.0

Load=200[lb/ft2] Sand 10 ft H*C σfinal ( ) ρc= 1+e σinitial log γT=100 [lb/ft3] ‘ ‘ Clay 40 ft and we only need to know it’s compression index, which we’ve been given. wc = 37% Cc=0.72 OCR=1.0

Load=200[lb/ft2] Sand 10 ft H*C σfinal ( ) ρc= 1+e σinitial log γT=100 [lb/ft3] ‘ ‘ Clay 40 ft We assume no consolidation settlement occurs in course grain soils, such as sand. wc = 37% Cc=0.72 OCR=1.0

Load=200[lb/ft2] Sand 10 ft H*C σfinal ( ) ρc= 1+e σinitial log γT=100 [lb/ft3] ‘ ‘ Clay 40 ft So the only consolidation will occur in the clay layer. wc = 37% Cc=0.72 OCR=1.0

Load=200[lb/ft2] Sand 10 ft H*C σfinal ( ) ρc= 1+e σinitial log γT=100 [lb/ft3] ‘ ‘ Clay 40 ft The problem states to evaluate the consolidation in 4 layers, wc = 37% Cc=0.72 OCR=1.0

Load=200[lb/ft2] Sand 10 ft H*C σfinal ( ) ρc= 1+e σinitial log γT=100 [lb/ft3] ‘ ‘ Clay layer 1 40 ft layer 2 so, we’ll evaluate 4, equally thick, 10 foot layers of clay. wc = 37% Cc=0.72 OCR=1.0 layer 3 layer 4

Load=200[lb/ft2] Sand 10 ft H*C σfinal ( ) ρc= 1+e σinitial log γT=100 [lb/ft3] ‘ ‘ Clay layer 1 40 ft layer 2 Rather than evaluate the settlement from one layer, wc = 37% Cc=0.72 OCR=1.0 layer 3 layer 4

( ) Find: ρc [in] from load (4 layers) H*C σfinal ρcn= log
Load=200[lb/ft2] Sand 10 ft H*C σfinal ( ) γT=100 [lb/ft3] ‘ ρcn= log 1+e σinitial ‘ Clay layer 1 40 ft layer 2 we’ll evaluate the settlement from n separate layers, wc = 37% Cc=0.72 OCR=1.0 layer 3 layer 4

Load=200[lb/ft2] Sand 10 ft H*C σfinal ( ) γT=100 [lb/ft3] ‘ ρcn= log 1+e σinitial ‘ Clay layer 1 40 ft layer 2 where in this case n, equals 4, one for each layer. wc = 37% Cc=0.72 OCR=1.0 layer 3 ρc = ρc1+ρc2+ρc3+ρc4 layer 4

Load=200[lb/ft2] 10 [ft] Sand 10 ft H*C σfinal ( ) γT=100 [lb/ft3] ‘ ρcn= log 1+e σinitial ‘ Clay layer 1 40 ft layer 2 The height of each later is 10 feet, wc = 37% Cc=0.72 OCR=1.0 layer 3 ρc = ρc1+ρc2+ρc3+ρc4 layer 4

Load=200[lb/ft2] 10 [ft] 0.72 Sand 10 ft H*C σfinal ( ) γT=100 [lb/ft3] ‘ ρcn= log 1+e σinitial ‘ Clay layer 1 40 ft layer 2 and the compression index is 0.72, wc = 37% Cc=0.72 OCR=1.0 layer 3 ρc = ρc1+ρc2+ρc3+ρc4 layer 4

( ) ? Find: ρc [in] from load (4 layers) H*C σfinal ρcn= log
Load=200[lb/ft2] 10 [ft] 0.72 Sand 10 ft H*C σfinal ( ) γT=100 [lb/ft3] ‘ ρcn= log 1+e σinitial ‘ Clay layer 1 ? 40 ft layer 2 but we don’t know the initial void ratio of the clay layer. wc = 37% Cc=0.72 OCR=1.0 layer 3 ρc = ρc1+ρc2+ρc3+ρc4 layer 4

( ) ? ? Find: ρc [in] from load (4 layers) H*C σfinal ρcn= log
Load=200[lb/ft2] 10 [ft] ? 0.72 Sand 10 ft H*C σfinal ( ) γT=100 [lb/ft3] ‘ ρcn= log 1+e σinitial ‘ Clay layer 1 ? 40 ft layer 2 nor do we know the initial or final vertical effective stresses. wc = 37% Cc=0.72 OCR=1.0 layer 3 ρc = ρc1+ρc2+ρc3+ρc4 layer 4

Load=200[lb/ft2] Sand 10 ft H*C σfinal ( ) γT=100 [lb/ft3] ‘ ρcn= log 1+e σinitial ‘ Clay layer 1 40 ft layer 2 Let’s first solve for the initial void ratio. wc = 37% Cc=0.72 OCR=1.0 layer 3 ρc = ρc1+ρc2+ρc3+ρc4 layer 4

Load=200[lb/ft2] Sand 10 ft γT=100 [lb/ft3] Clay layer 1 40 ft layer 2 One way to solve this problem --- wc = 37% Cc=0.72 OCR=1.0 layer 3 layer 4

γd= Find: ρc [in] from load (4 layers) γT=100 [lb/ft3] SG * γw Sand
Load=200[lb/ft2] γd= SG * γw wc*SG+1 Sand 10 ft γT=100 [lb/ft3] Clay layer 1 40 ft layer 2 is to first solve for the dry unit weight, wc = 37% Cc=0.72 OCR=1.0 layer 3 layer 4

Load=200[lb/ft2] γd= SG * γw wc*SG+1 2.70 Sand 10 ft γT=100 [lb/ft3] Clay layer 1 40 ft layer 2 where, the specific gravity of clay is 2.7. wc = 37% Cc=0.72 OCR=1.0 layer 3 layer 4

Load=200[lb/ft2] γd= SG * γw wc*SG+1 2.70 Sand 10 ft γT=100 [lb/ft3] 62.4 [lb/ft3] Clay layer 1 40 ft layer 2 The unit weight of water is 62.4 pounds per cubic feet, wc = 37% Cc=0.72 OCR=1.0 layer 3 layer 4

Load=200[lb/ft2] γd= SG * γw wc*SG+1 2.70 Sand 10 ft γT=100 [lb/ft3] 62.4 [lb/ft3] Clay layer 1 0.37 40 ft layer 2 And the water content is 0.37. wc = 37% Cc=0.72 OCR=1.0 layer 3 layer 4

γd= γd = 84.28 [lb/ft3] Find: ρc [in] from load (4 layers)
Load=200[lb/ft2] γd= SG * γw wc*SG+1 2.70 Sand 10 ft γT=100 [lb/ft3] 62.4 [lb/ft3] Clay layer 1 0.37 γd = [lb/ft3] 40 ft layer 2 The dry unit weight is pounds per cubic foot. Next we solve --- wc = 37% Cc=0.72 OCR=1.0 layer 3 layer 4

γd= γd = 84.28 [lb/ft3] γd Find: ρc [in] from load (4 layers) e= -1
Load=200[lb/ft2] γd= SG * γw wc*SG+1 2.70 Sand 10 ft γT=100 [lb/ft3] 62.4 [lb/ft3] Clay layer 1 0.37 γd = [lb/ft3] 40 ft layer 2 for the void ratio and plug in our wc = 37% Cc=0.72 OCR=1.0 layer 3 e= γd SG*γw layer 4

Load=200[lb/ft2] γd= SG * γw wc*SG+1 2.70 Sand 10 ft γT=100 [lb/ft3] 62.4 [lb/ft3] Clay layer 1 0.37 γd = [lb/ft3] 40 ft layer 2 specific gravity, wc = 37% Cc=0.72 OCR=1.0 layer 3 e= γd SG*γw layer 4

Load=200[lb/ft2] γd= SG * γw wc*SG+1 2.70 Sand 10 ft γT=100 [lb/ft3] 62.4 [lb/ft3] Clay layer 1 0.37 γd = [lb/ft3] 40 ft layer 2 unit weight of water, wc = 37% Cc=0.72 OCR=1.0 layer 3 e= γd SG*γw layer 4

Load=200[lb/ft2] γd= SG * γw wc*SG+1 2.70 Sand 10 ft γT=100 [lb/ft3] 62.4 [lb/ft3] Clay layer 1 0.37 γd = [lb/ft3] 40 ft layer 2 and dry unit weight. wc = 37% Cc=0.72 OCR=1.0 layer 3 e= γd SG*γw layer 4

Load=200[lb/ft2] γd= SG * γw wc*SG+1 2.70 Sand 10 ft γT=100 [lb/ft3] 62.4 [lb/ft3] Clay layer 1 0.37 γd = [lb/ft3] 40 ft layer 2 The initial void ratio of the clay is 1. wc = 37% Cc=0.72 OCR=1.0 layer 3 e= γd SG*γw layer 4 e=1.00

Load=200[lb/ft2] 10 [ft] 0.72 Sand 10 ft H*C σfinal ( ) γT=100 [lb/ft3] ‘ ρcn= log 1+e σinitial ‘ Clay layer 1 ? 40 ft layer 2 --- wc = 37% Cc=0.72 OCR=1.0 layer 3 ρc = ρc1+ρc2+ρc3+ρc4 layer 4

Load=200[lb/ft2] 10 [ft] 0.72 Sand 10 ft H*C σfinal ( ) γT=100 [lb/ft3] ‘ ρcn= log 1+e σinitial ‘ Clay layer 1 40 ft layer 2 Next we solve for --- 1.00 wc = 37% Cc=0.72 OCR=1.0 layer 3 ρc = ρc1+ρc2+ρc3+ρc4 layer 4

Load=200[lb/ft2] 10 [ft] ? 0.72 Sand 10 ft H*C σfinal ( ) γT=100 [lb/ft3] ‘ ρcn= log 1+e σinitial ‘ Clay layer 1 40 ft layer 2 the vertical effective stresses of the clay layer both before and after loading. 1.00 wc = 37% Cc=0.72 OCR=1.0 layer 3 ρc = ρc1+ρc2+ρc3+ρc4 layer 4

Load=200[lb/ft2] 10 [ft] ? 0.72 Sand 10 ft H*C σfinal ( ) γT=100 [lb/ft3] ‘ ρcn= log 1+e σinitial ‘ Clay layer 1 40 ft layer 2 Let’s first work through layer 1 --- 1.00 wc = 37% Cc=0.72 OCR=1.0 layer 3 ρc = ρc1+ρc2+ρc3+ρc4 layer 4

Load=200[lb/ft2] 10 [ft] ? 0.72 Sand 10 ft H*C σfinal ( ) γT=100 [lb/ft3] ‘ ρcn= log 1+e σinitial ‘ Clay layer 1 40 ft layer 2 which is the first --- 1.00 wc = 37% Cc=0.72 OCR=1.0 layer 3 ρc = ρc1+ρc2+ρc3+ρc4 layer 4

Load=200[lb/ft2] 10 [ft] ? 0.72 Sand 10 ft H*C σfinal ( ) γT=100 [lb/ft3] ‘ ρcn= log 1+e σinitial ‘ Clay layer 1 40 ft layer 2 Of 4 layers. 1.00 wc = 37% Cc=0.72 OCR=1.0 layer 3 ρc = ρc1+ρc2+ρc3+ρc4 layer 4

( ) Find: ρc [in] from load (4 layers) H*C σfinal ρc1= log
Load=200[lb/ft2] 10 [ft] 0.72 Sand 10 ft H*C σfinal ( ) γT=100 [lb/ft3] ‘ ρc1= log 1+e σinitial ‘ Clay layer 1 40 ft layer 2 We’ll solve for the initial --- 1.00 wc = 37% Cc=0.72 OCR=1.0 layer 3 layer 4

Load=200[lb/ft2] 10 [ft] 0.72 Sand 10 ft H*C σfinal ( ) γT=100 [lb/ft3] ‘ ρc1= log 1+e σinitial ‘ Clay layer 1 40 ft layer 2 vertical effective stress of layer 1 at a point 1.00 wc = 37% Cc=0.72 OCR=1.0 layer 3 layer 4

Load=200[lb/ft2] 10 [ft] 0.72 Sand 10 ft H*C σfinal ( ) γT=100 [lb/ft3] ‘ ρc1= log 1+e σinitial ‘ Clay layer 1 40 ft layer 2 in the middle of the layer, which in this case --- 1.00 wc = 37% Cc=0.72 OCR=1.0 layer 3 layer 4

Load=200[lb/ft2] 10 [ft] 0.72 Sand 10 ft H*C σfinal ( ) γT=100 [lb/ft3] ‘ ρc1= log 1+e σinitial ‘ Clay layer 1 40 ft layer 2 would be 5 feet beneath the sand-clay interface. 1.00 wc = 37% Cc=0.72 OCR=1.0 layer 3 layer 4

Load=200[lb/ft2] 10 [ft] 0.72 Sand 10 ft H*C σfinal ( ) γT=100 [lb/ft3] ‘ ρc1= log 1+e σinitial ‘ Clay layer 1 40 ft layer 2 So we multiply the effective unit weight times the height, 1.00 wc = 37% Cc=0.72 OCR=1.0 σ’1,initial = Σ γ’ * d layer 3 layer 4

Load=200[lb/ft2] 10 [ft] 0.72 Sand 10 ft H*C σfinal ( ) γT=100 [lb/ft3] ‘ ρc1= log 1+e σinitial ‘ Clay layer 1 40 ft layer 2 for both the sand layer and the portion of the clay layer above our point of interest. 1.00 wc = 37% Cc=0.72 OCR=1.0 σ’1,initial = Σ γ’ * d layer 3 σ’1,initial = γ’1,sand*d1,sand layer 4 + γ’1,clay*d1,clay

Load=200[lb/ft2] Sand 10 ft γT=100 [lb/ft3] Clay layer 1 40 ft layer 2 wc = 37% Cc=0.72 OCR=1.0 σ’1,initial = Σ γ’ * d layer 3 σ’1,initial = γ’1,sand*d1,sand layer 4 + γ’1,clay*d1,clay

Load=200[lb/ft2] σ’1,initial = Σ γ * d Sand 10 ft σ’1,initial = γ’sand*dsand γT=100 [lb/ft3] + γ’1,clay*d1,clay Clay layer 1 40 ft layer 2 The effective unit weight of sand --- wc = 37% Cc=0.72 OCR=1.0 layer 3 layer 4

Load=200[lb/ft2] σ’1,initial = Σ γ * d Sand 100 [lb/ft3] 10 ft σ’1,initial = γ’sand*dsand γT=100 [lb/ft3] + γ’1,clay*d1,clay Clay layer 1 40 ft layer 2 is 100 pounds per cubic foot, which is the same as the total unit weight, since there is no pore water pressure. wc = 37% Cc=0.72 OCR=1.0 layer 3 layer 4

Load=200[lb/ft2] σ’1,initial = Σ γ * d 10 [ft] Sand 100 [lb/ft3] 10 ft σ’1,initial = γ’sand*dsand γT=100 [lb/ft3] + γ’1,clay*d1,clay Clay layer 1 40 ft layer 2 The depth of the sand layer is 10 feet. The depth of the clay layer --- wc = 37% Cc=0.72 OCR=1.0 layer 3 layer 4

Load=200[lb/ft2] σ’1,initial = Σ γ * d 10 [ft] Sand 100 [lb/ft3] 10 ft σ’1,initial = γ’sand*dsand γT=100 [lb/ft3] + γ’1,clay*d1,clay Clay layer 1 40 ft layer 2 In the middle of layer 1, is 5 feet, 5 [ft] wc = 37% Cc=0.72 OCR=1.0 layer 3 layer 4

? Find: ρc [in] from load (4 layers) σ’1,initial = Σ γ * d
Load=200[lb/ft2] σ’1,initial = Σ γ * d 10 [ft] Sand 100 [lb/ft3] 10 ft σ’1,initial = γ’sand*dsand γT=100 [lb/ft3] + γ’1,clay*d1,clay Clay layer 1 ? 40 ft layer 2 And we do not know the effective unit weight of the clay layer. 5 [ft] wc = 37% Cc=0.72 OCR=1.0 layer 3 layer 4

? Find: ρc [in] from load (4 layers) γd = 84.28 [lb/ft3]
Load=200[lb/ft2] σ’1,initial = Σ γ * d 10 [ft] Sand 100 [lb/ft3] 10 ft σ’1,initial = γ’sand*dsand γT=100 [lb/ft3] + γ’1,clay*d1,clay Clay layer 1 ? 40 ft layer 2 However, when solving for the initial void ratio, in one of our intermediate steps, we determined the dry unit weight. 5 [ft] γd = [lb/ft3] wc = 37% Cc=0.72 OCR=1.0 layer 3 layer 4

Load=200[lb/ft2] σ’1,initial = Σ γ * d 10 [ft] Sand 100 [lb/ft3] 10 ft σ’1,initial = γ’sand*dsand γT=100 [lb/ft3] + γ’1,clay*d1,clay Clay layer 1 ? 40 ft layer 2 And knowing the water content, we can solve for --- 5 [ft] γd = [lb/ft3] wc = 37% Cc=0.72 OCR=1.0 layer 3 wc=37% layer 4

Load=200[lb/ft2] σ’1,initial = Σ γ * d 10 [ft] Sand 100 [lb/ft3] 10 ft σ’1,initial = γ’sand*dsand γT=100 [lb/ft3] + γ’1,clay*d1,clay Clay layer 1 ? 40 ft layer 2 the total unit weight 5 [ft] γd = [lb/ft3] wc = 37% Cc=0.72 OCR=1.0 layer 3 wc=37% layer 4 γT = γd * (1+wc)

Load=200[lb/ft2] σ’1,initial = Σ γ * d 10 [ft] Sand 100 [lb/ft3] 10 ft σ’1,initial = γ’sand*dsand γT=100 [lb/ft3] + γ’1,clay*d1,clay Clay layer 1 ? 40 ft layer 2 We plug in our values 5 [ft] γd = [lb/ft3] wc = 37% Cc=0.72 OCR=1.0 layer 3 wc=37% layer 4 γT = γd * (1+wc)

Load=200[lb/ft2] σ’1,initial = Σ γ * d 10 [ft] Sand 100 [lb/ft3] 10 ft σ’1,initial = γ’sand*dsand γT=100 [lb/ft3] + γ’1,clay*d1,clay Clay layer 1 ? 40 ft layer 2 And the total unit weight of the clay layer is --- 5 [ft] γd = [lb/ft3] wc = 37% Cc=0.72 OCR=1.0 layer 3 wc=37%=0.37 layer 4 γT = γd * (1+wc)

Load=200[lb/ft2] σ’1,initial = Σ γ * d 10 [ft] Sand 100 [lb/ft3] 10 ft σ’1,initial = γ’sand*dsand γT=100 [lb/ft3] + γ’1,clay*d1,clay Clay layer 1 ? 40 ft layer 2 pounds per cubic feet. 5 [ft] γd = [lb/ft3] wc = 37% Cc=0.72 OCR=1.0 layer 3 wc=37%=0.37 layer 4 γT = γd * (1+wc) γT = [lb/ft3]

Load=200[lb/ft2] σ’1,initial = Σ γ * d 10 [ft] Sand 100 [lb/ft3] 10 ft σ’1,initial = γ’sand*dsand γT=100 [lb/ft3] + γ’1,clay*d1,clay Clay layer 1 ? 40 ft layer 2 So we subtract --- 5 [ft] γd = [lb/ft3] wc = 37% Cc=0.72 OCR=1.0 layer 3 wc=37%=0.37 layer 4 γT = γd * (1+wc) γ’1,clay = γT - γw γT = [lb/ft3]

Load=200[lb/ft2] σ’1,initial = Σ γ * d 10 [ft] Sand 100 [lb/ft3] 10 ft σ’1,initial = γ’sand*dsand γT=100 [lb/ft3] + γ’1,clay*d1,clay Clay layer 1 ? 40 ft layer 2 minus --- 5 [ft] γd = [lb/ft3] wc = 37% Cc=0.72 OCR=1.0 layer 3 wc=37%=0.37 layer 4 γT = γd * (1+wc) γ’1,clay = γT - γw γT = [lb/ft3]

Load=200[lb/ft2] σ’1,initial = Σ γ * d 10 [ft] Sand 100 [lb/ft3] 10 ft σ’1,initial = γ’sand*dsand γT=100 [lb/ft3] + γ’1,clay*d1,clay Clay layer 1 40 ft layer 2 62.4, to get an effective unit weight of 5 [ft] γd = [lb/ft3] wc = 37% Cc=0.72 OCR=1.0 layer 3 wc=37%=0.37 layer 4 γT = γd * (1+wc) γ’1,clay = γT - γw γT = [lb/ft3] 62.4 [lb/ft3]

Load=200[lb/ft2] σ’1,initial = Σ γ * d 10 [ft] Sand 100 [lb/ft3] 10 ft σ’1,initial = γ’sand*dsand γT=100 [lb/ft3] + γ’1,clay*d1,clay Clay layer 1 40 ft layer 2 53.1 [lb/ft3] 53.1 pounds per cubic feet, in the clay layer. 5 [ft] γd = [lb/ft3] wc = 37% Cc=0.72 OCR=1.0 layer 3 wc=37%=0.37 layer 4 γT = γd * (1+wc) γ’1,clay = γT - γw γT = [lb/ft3] 62.4 [lb/ft3]

Load=200[lb/ft2] σ’1,initial = Σ γ * d 10 [ft] Sand 100 [lb/ft3] 10 ft σ’1,initial = γ’sand*dsand γT=100 [lb/ft3] + γ’1,clay*d1,clay Clay layer 1 40 ft layer 2 53.1 [lb/ft3] So our initial vertical effective stress, 5 [ft] wc = 37% Cc=0.72 OCR=1.0 layer 3 layer 4

Load=200[lb/ft2] σ’1,initial = Σ γ * d 10 [ft] Sand 100 [lb/ft3] 10 ft σ’1,initial = γ’sand*dsand γT=100 [lb/ft3] + γ’1,clay*d1,clay Clay layer 1 40 ft layer 2 53.1 [lb/ft3] for layer 1 is 1,266 pounds per square feet. The final vertical effective stress --- 5 [ft] wc = 37% Cc=0.72 OCR=1.0 σ’1,initial = 1,266 [lb/ft2] layer 3 layer 4

Load=200[lb/ft2] σ’1,initial = Σ γ * d 10 [ft] Sand 100 [lb/ft3] 10 ft σ’1,initial = γ’sand*dsand γT=100 [lb/ft3] + γ’1,clay*d1,clay Clay layer 1 40 ft layer 2 53.1 [lb/ft3] is equal to the initial vertical --- 5 [ft] wc = 37% Cc=0.72 OCR=1.0 σ’1,initial = 1,266 [lb/ft2] layer 3 σ’1,final = σ’1,initial +load layer 4

Load=200[lb/ft2] σ’1,initial = Σ γ * d 10 [ft] Sand 100 [lb/ft3] 10 ft σ’1,initial = γ’sand*dsand γT=100 [lb/ft3] + γ’1,clay*d1,clay Clay layer 1 40 ft layer 2 53.1 [lb/ft3] effective stress, 5 [ft] wc = 37% Cc=0.72 OCR=1.0 σ’1,initial = 1,266 [lb/ft2] layer 3 σ’1,final = σ’1,initial +load layer 4

Load=200[lb/ft2] σ’1,initial = Σ γ * d 10 [ft] Sand 100 [lb/ft3] 10 ft σ’1,initial = γ’sand*dsand γT=100 [lb/ft3] + γ’1,clay*d1,clay Clay layer 1 40 ft layer 2 53.1 [lb/ft3] plus the added load, of 200 pounds per square foot 5 [ft] wc = 37% Cc=0.72 OCR=1.0 σ’1,initial = 1,266 [lb/ft2] layer 3 σ’1,final = σ’1,initial +load layer 4

Load=200[lb/ft2] σ’1,initial = Σ γ * d 10 [ft] Sand 100 [lb/ft3] 10 ft σ’1,initial = γ’sand*dsand γT=100 [lb/ft3] + γ’1,clay*d1,clay Clay layer 1 40 ft layer 2 53.1 [lb/ft3] Or, 1,466 pounds per square foot. 5 [ft] wc = 37% Cc=0.72 OCR=1.0 σ’1,initial = 1,266 [lb/ft2] layer 3 σ’1,final = σ’1,initial +load layer 4 σ’1,final = 1,466 [lb/ft2]

Load=200[lb/ft2] σ’1,initial = Σ γ * d 10 [ft] Sand 100 [lb/ft3] 10 ft σ’1,initial = γ’sand*dsand γT=100 [lb/ft3] + γ’1,clay*d1,clay Clay layer 1 40 ft layer 2 53.1 [lb/ft3] --- 5 [ft] wc = 37% Cc=0.72 OCR=1.0 σ’1,initial = 1,266 [lb/ft2] layer 3 σ’1,final = σ’1,initial +load layer 4 σ’1,final = 1,466 [lb/ft2]

Load=200[lb/ft2] σ’1,initial = Σ γ * d 10 [ft] Sand 100 [lb/ft3] 10 ft σ’1,initial = γ’sand*dsand γT=100 [lb/ft3] + γ’1,clay*d1,clay Clay layer 1 40 ft layer 2 53.1 [lb/ft3] So we have our initial --- 5 [ft] wc = 37% Cc=0.72 OCR=1.0 σ’1,initial = 1,266 [lb/ft2] layer 3 σ’1,final = σ’1,initial +load layer 4 σ’1,final = 1,466 [lb/ft2]

Load=200[lb/ft2] σ’1,initial = Σ γ * d 10 [ft] Sand 100 [lb/ft3] 10 ft σ’1,initial = γ’sand*dsand γT=100 [lb/ft3] + γ’1,clay*d1,clay Clay layer 1 40 ft layer 2 53.1 [lb/ft3] and final stress values --- 5 [ft] wc = 37% Cc=0.72 OCR=1.0 σ’1,initial = 1,266 [lb/ft2] layer 3 σ’1,final = σ’1,initial +load layer 4 σ’1,final = 1,266 [lb/ft2]

Load=200[lb/ft2] Sand 10 ft γT=100 [lb/ft3] Clay layer 1 40 ft layer 2 and can solve for the--- wc = 37% Cc=0.72 OCR=1.0 layer 3 σ’1,initial = 1,266 [lb/ft2] layer 4 σ’1,final = 1,466 [lb/ft2]

Load=200[lb/ft2] 10 [ft] 0.72 Sand 10 ft H*C σfinal ( ) γT=100 [lb/ft3] ‘ ρc1= log 1+e σinitial ‘ Clay layer 1 1.00 40 ft layer 2 consolidation settlement --- wc = 37% Cc=0.72 OCR=1.0 layer 3 σ’1,initial = 1,266 [lb/ft2] layer 4 σ’1,final = 1,466 [lb/ft2]

Load=200[lb/ft2] 10 [ft] 0.72 Sand 10 ft H*C σfinal ( ) γT=100 [lb/ft3] ‘ ρc1= log 1+e σinitial ‘ Clay layer 1 1.00 40 ft layer 2 of the first layer, 1,266 [lb/ft2] wc = 37% Cc=0.72 OCR=1.0 layer 3 σ’1,initial = 1,266 [lb/ft2] layer 4 σ’1,final = 1,466 [lb/ft2]

Load=200[lb/ft2] 10 [ft] 0.72 1,466 [lb/ft2] Sand 10 ft H*C σfinal ( ) γT=100 [lb/ft3] ‘ ρc1= log 1+e σinitial ‘ Clay layer 1 1.00 40 ft layer 2 Which turns out to be, 1,266 [lb/ft2] wc = 37% Cc=0.72 OCR=1.0 layer 3 σ’1,initial = 1,266 [lb/ft2] layer 4 σ’1,final = 1,466 [lb/ft2]

Load=200[lb/ft2] 10 [ft] 0.72 1,466 [lb/ft2] Sand 10 ft H*C σfinal ( ) γT=100 [lb/ft3] ‘ ρc1= log 1+e σinitial ‘ Clay layer 1 1.00 40 ft layer 2 2.75 inches. 1,266 [lb/ft2] wc = 37% Cc=0.72 OCR=1.0 layer 3 ρc1=2.75 [in] layer 4

Load=200[lb/ft2] 10 [ft] 0.72 1,466 [lb/ft2] Sand 10 ft H*C σfinal ( ) γT=100 [lb/ft3] ‘ ρc1= log 1+e σinitial ‘ Clay layer 1 1.00 40 ft layer 2 Keep in mind, 1,266 [lb/ft2] wc = 37% Cc=0.72 OCR=1.0 layer 3 ρc1=2.75 [in] layer 4 ρc = ρc1+ρc2+ρc3+ρc4

Load=200[lb/ft2] 10 [ft] 0.72 1,466 [lb/ft2] Sand 10 ft H*C σfinal ( ) γT=100 [lb/ft3] ‘ ρc1= log 1+e σinitial ‘ Clay layer 1 1.00 40 ft layer 2 this is just one of the four, 1,266 [lb/ft2] wc = 37% Cc=0.72 OCR=1.0 layer 3 ρc1=2.75 [in] layer 4 ρc = ρc1+ρc2+ρc3+ρc4

Load=200[lb/ft2] 10 [ft] 0.72 1,466 [lb/ft2] Sand 10 ft H*C σfinal ( ) γT=100 [lb/ft3] ‘ ρc1= log 1+e σinitial ‘ Clay layer 1 1.00 40 ft layer 2 10-foot thick clay layers to evaluate. 1,266 [lb/ft2] wc = 37% Cc=0.72 OCR=1.0 layer 3 ρc1=2.75 [in] layer 4 ρc = ρc1+ρc2+ρc3+ρc4

Load=200[lb/ft2] 10 [ft] 0.72 1,466 [lb/ft2] Sand 10 ft H*C σfinal ( ) γT=100 [lb/ft3] ‘ ρc1= log 1+e σinitial ‘ Clay layer 1 1.00 40 ft layer 2 Layers 2, 3 and 4 can be solved in a similar fashion. 1,266 [lb/ft2] wc = 37% Cc=0.72 OCR=1.0 layer 3 ρc1=2.75 [in] layer 4 ρc = ρc1+ρc2+ρc3+ρc4

Load=200[lb/ft2] 10 [ft] 0.72 Sand 10 ft H*C σfinal ( ) γT=100 [lb/ft3] ‘ ρcn= log 1+e σinitial ‘ Clay layer 1 1.00 40 ft layer 2 The first part of the settlement equation is the same, wc = 37% Cc=0.72 OCR=1.0 layer 3 ρc1=2.75 [in] layer 4 ρc = ρc1+ρc2+ρc3+ρc4

Load=200[lb/ft2] 10 [ft] 0.72 Sand 10 ft H*C σfinal ( ) γT=100 [lb/ft3] ‘ ρcn= log 1+e σinitial ‘ Clay layer 1 1.00 40 ft layer 2 The only difference is the initial and final stress values. wc = 37% Cc=0.72 OCR=1.0 layer 3 ρc1=2.75 [in] layer 4 ρc = ρc1+ρc2+ρc3+ρc4

Load=200[lb/ft2] σ’initial = γ’sand*dsand Sand 10 ft γT=100 [lb/ft3] + γ’clay*dclay Clay layer 1 40 ft layer 2 The initial effective vertical stress -- wc = 37% Cc=0.72 OCR=1.0 layer 3 layer 4

Load=200[lb/ft2] 100 [lb/ft3] 10 [ft] σ’initial = γ’sand*dsand Sand 10 ft γT=100 [lb/ft3] + γ’clay*dclay Clay layer 1 53.1 [lb/ft3] 40 ft layer 2 for layers 2, 3 and 4. Will be evaluated at points --- wc = 37% Cc=0.72 OCR=1.0 layer 3 layer 4

Load=200[lb/ft2] 100 [lb/ft3] 10 [ft] σ’initial = γ’sand*dsand Sand 10 ft γT=100 [lb/ft3] + γ’clay*dclay Clay layer 1 53.1 [lb/ft3] 15 [ft] 40 ft layer 2 15, wc = 37% Cc=0.72 OCR=1.0 layer 3 layer 4

Load=200[lb/ft2] 100 [lb/ft3] 10 [ft] σ’initial = γ’sand*dsand Sand 10 ft γT=100 [lb/ft3] + γ’clay*dclay Clay layer 1 53.1 [lb/ft3] 15 [ft] 40 ft layer 2 25 [ft] 25, wc = 37% Cc=0.72 OCR=1.0 layer 3 layer 4

Load=200[lb/ft2] 100 [lb/ft3] 10 [ft] σ’initial = γ’sand*dsand Sand 10 ft γT=100 [lb/ft3] + γ’clay*dclay Clay layer 1 53.1 [lb/ft3] 15 [ft] 40 ft layer 2 25 [ft] And 35 feet beneath the sand-clay interface. At this point it would be --- 35 [ft] wc = 37% Cc=0.72 OCR=1.0 layer 3 layer 4

Load=200[lb/ft2] 100 [lb/ft3] 10 [ft] σ’initial = γ’sand*dsand Sand 10 ft γT=100 [lb/ft3] + γ’clay*dclay Clay layer 1 53.1 [lb/ft3] 15 [ft] 40 ft layer 2 25 [ft] convenient to set up a table of values --- 35 [ft] wc = 37% Cc=0.72 OCR=1.0 Layer σ’initial σ’final layer 3 2 layer 4 3 4

Load=200[lb/ft2] 100 [lb/ft3] 10 [ft] σ’initial = γ’sand*dsand Sand 10 ft γT=100 [lb/ft3] + γ’clay*dclay Clay layer 1 53.1 [lb/ft3] 15 [ft] 40 ft layer 2 25 [ft] For the initial and final --- 35 [ft] wc = 37% Cc=0.72 OCR=1.0 Layer σ’initial σ’final layer 3 ,797 layer 4 ,328 ,859

Load=200[lb/ft2] 100 [lb/ft3] 10 [ft] σ’initial = γ’sand*dsand Sand 10 ft γT=100 [lb/ft3] + γ’clay*dclay Clay layer 1 53.1 [lb/ft3] 15 [ft] 40 ft layer 2 25 [ft] effective vertical stresses --- 35 [ft] wc = 37% Cc=0.72 OCR=1.0 Layer σ’initial σ’final layer 3 ,797 layer 4 ,328 ,859 σ’final = σ’initial +load

Load=200[lb/ft2] 100 [lb/ft3] 10 [ft] σ’initial = γ’sand*dsand Sand 10 ft γT=100 [lb/ft3] + γ’clay*dclay Clay layer 1 53.1 [lb/ft3] 15 [ft] 40 ft layer 2 25 [ft] where the final is equal to the initial 35 [ft] wc = 37% Cc=0.72 OCR=1.0 Layer σ’initial σ’final layer 3 ,797 layer 4 ,328 ,859 σ’final = σ’initial + load

Load=200[lb/ft2] 100 [lb/ft3] 10 [ft] σ’initial = γ’sand*dsand Sand 10 ft γT=100 [lb/ft3] + γ’clay*dclay Clay layer 1 53.1 [lb/ft3] 15 [ft] 40 ft layer 2 25 [ft] plus the load. 35 [ft] wc = 37% Cc=0.72 OCR=1.0 Layer σ’initial σ’final layer 3 , ,997 layer 4 , ,528 , ,059 σ’final = σ’initial + load

Load=200[lb/ft2] 100 [lb/ft3] 10 [ft] σ’initial = γ’sand*dsand Sand 10 ft γT=100 [lb/ft3] + γ’clay*dclay Clay layer 1 53.1 [lb/ft3] 15 [ft] 40 ft layer 2 25 [ft] Returning to our --- 35 [ft] wc = 37% Cc=0.72 OCR=1.0 Layer σ’initial σ’final layer 3 , ,997 layer 4 , ,528 , ,059 σ’final = σ’initial + load

Load=200[lb/ft2] ( ) H*C σfinal ρcn= 1+e σinitial log ‘ Sand 10 ft γT=100 [lb/ft3] Clay layer 1 40 ft layer 2 Consolidation settlement equation, wc = 37% Cc=0.72 OCR=1.0 layer 3 layer 4

Load=200[lb/ft2] 10[ft]*0.72 ρc2= 1,997 Sand log 10 ft 1+1 1,797 γT=100 [lb/ft3] Clay layer 1 40 ft layer 2 The second layer will consolidate --- wc = 37% Cc=0.72 OCR=1.0 layer 3 layer 4

Load=200[lb/ft2] 10[ft]*0.72 ρc2= 1,997 Sand log 10 ft 1+1 1,797 γT=100 [lb/ft3] ρc2= 1.98[in] Clay layer 1 40 ft layer 2 1.98 inches, wc = 37% Cc=0.72 OCR=1.0 layer 3 layer 4

Load=200[lb/ft2] 10[ft]*0.72 ρc2= 1,997 Sand log 10 ft 1+1 1,797 γT=100 [lb/ft3] ρc2= 1.98[in] Clay layer 1 10[ft]*0.72 ρc3= 2,528 log 40 ft layer 2 The third layer will consolidate --- 1+1 2,328 wc = 37% Cc=0.72 OCR=1.0 layer 3 layer 4

Load=200[lb/ft2] 10[ft]*0.72 ρc2= 1,997 Sand log 10 ft 1+1 1,797 γT=100 [lb/ft3] ρc2= 1.98[in] Clay layer 1 10[ft]*0.72 ρc3= 2,528 log 40 ft layer 2 1.62 inches. And the 1+1 2,328 wc = 37% Cc=0.72 OCR=1.0 layer 3 ρc3= 1.62[in] layer 4

Load=200[lb/ft2] 10[ft]*0.72 ρc2= 1,997 Sand log 10 ft 1+1 1,797 γT=100 [lb/ft3] ρc2= 1.98[in] Clay layer 1 10[ft]*0.72 ρc3= 2,528 log 40 ft layer 2 Fourth layer will consolidate 1+1 2,328 wc = 37% Cc=0.72 OCR=1.0 layer 3 ρc3= 1.62[in] 10[ft]*0.72 layer 4 ρc4= 3,059 log 1+1 2,859

Load=200[lb/ft2] 10[ft]*0.72 ρc2= 1,997 Sand log 10 ft 1+1 1,797 γT=100 [lb/ft3] ρc2= 1.98[in] Clay layer 1 10[ft]*0.72 ρc3= 2,528 log 40 ft layer 2 1.27 inches. 1+1 2,328 wc = 37% Cc=0.72 OCR=1.0 layer 3 ρc3= 1.62[in] 10[ft]*0.72 layer 4 ρc4= 3,059 log 1+1 2,859 = 1.27[in]

Load=200[lb/ft2] Sand 10 ft γT=100 [lb/ft3] ρc = ρc1+ρc2+ρc3+ρc4 Clay layer 1 40 ft layer 2 Adding the settlement from all 4 layers, wc = 37% Cc=0.72 OCR=1.0 layer 3 layer 4

Load=200[lb/ft2] 2.75 [in] 1.62 [in] Sand 10 ft γT=100 [lb/ft3] ρc = ρc1+ρc2+ρc3+ρc4 Clay layer 1 1.27 [in] 1.98 [in] 40 ft layer 2 The settlement from consolidation is --- wc = 37% Cc=0.72 OCR=1.0 layer 3 layer 4

Load=200[lb/ft2] 2.75 [in] 1.62 [in] Sand 10 ft γT=100 [lb/ft3] ρc = ρc1+ρc2+ρc3+ρc4 Clay layer 1 1.27 [in] 1.98 [in] 40 ft layer 2 7.62 inches. When compared to --- wc = 37% Cc=0.72 OCR=1.0 ρc = 7.62 [in] layer 3 layer 4

1.0 2.5 C) 5 D) 7.5 Load=200[lb/ft2] 2.75 [in] 1.62 [in] Sand 10 ft γT=100 [lb/ft3] ρc = ρc1+ρc2+ρc3+ρc4 Clay layer 1 1.27 [in] 1.98 [in] 40 ft layer 2 the possible solutions, wc = 37% Cc=0.72 OCR=1.0 ρc = 7.62 [in] layer 3 layer 4

1.0 2.5 C) 5 D) 7.5 Load=200[lb/ft2] 2.75 [in] 1.62 [in] Sand 10 ft γT=100 [lb/ft3] ρc = ρc1+ρc2+ρc3+ρc4 Clay layer 1 1.27 [in] 1.98 [in] 40 ft layer 2 the answer is D. [end] wc = 37% Cc=0.72 OCR=1.0 ρc = 7.62 [in] layer 3 layer 4 Answer: D

( ) ? 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 ‘

Find: ρc [in] from load (4 layers)

Παρόμοιες παρουσιάσεις

Παρουσίαση με θέμα: "Find: ρc [in] from load (4 layers)"— Μεταγράφημα παρουσίασης:

Παρόμοιες παρουσιάσεις

Σχετικά με το έργο

Σχόλια

Είσοδος

Σύνδεση μέσω των κοινωνικών δικτύων:

Find: ρc [in] from load (4 layers)

Παρόμοιες παρουσιάσεις

Παρουσίαση με θέμα: "Find: ρc [in] from load (4 layers)"— Μεταγράφημα παρουσίασης:

Παρόμοιες παρουσιάσεις

Σχετικά με το έργο

Σχόλια