Find: CC Lab Test Data e C) 0.38 D) 0.50 Load [kPa] 0.919

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Find: CC Lab Test Data e 0.03 0.07 C) 0.38 D) 0.50 Load [kPa] 0.919 0.900 0.890 0.868 0.825 0.771 0.687 0.572 0.463 Load [kPa] 6.25 12.5 25 50 100 200 400 800 Find the compression index, also referred to as the virgin compression index, C sub c. In this problem,

Find: CC Lab Test Data e 0.03 0.07 C) 0.38 D) 0.50 Load [kPa] 0.919 0.900 0.890 0.868 0.825 0.771 0.687 0.572 0.463 Load [kPa] 6.25 12.5 25 50 100 200 400 800 We’ve been provided laboratory data from a consolidation test.

Find: CC Lab Test Data e Load [kPa] 0.919 6.25 0.900 0.890 12.5 25 0.868 0.825 0.771 0.687 0.572 0.463 Load [kPa] 6.25 12.5 25 50 100 200 400 800 The compression index can be determined by plotting ---

Find: CC e Lab Test Data e Load [kPa] Load [kPa] 0.919 6.25 0.900 0.890 0.868 0.825 0.771 0.687 0.572 0.463 Load [kPa] 6.25 12.5 25 50 100 200 400 800 the laboratory test data on a semi-log graph, where --- Load [kPa]

Find: CC e Lab Test Data e Load [kPa] Load [kPa] 0.919 6.25 0.900 0.890 0.868 0.825 0.771 0.687 0.572 0.463 Load [kPa] 6.25 12.5 25 50 100 200 400 800 the vertical axis is the void ratio, Load [kPa]

Find: CC e Lab Test Data e Load [kPa] Load [kPa] 0.919 6.25 0.900 0.890 0.868 0.825 0.771 0.687 0.572 0.463 Load [kPa] 6.25 12.5 25 50 100 200 400 800 And the horizontal axis is the load. This horizontal axis is plotted on a log scale, and the units of load for this problem, are in kilopascals. Load [kPa]

Find: CC e Lab Test Data e Load [kPa] 0.9 0.8 0.7 0.6 0.5 0.4 1 10 100 0.919 0.900 0.890 0.868 0.825 0.771 0.687 0.572 0.463 Load [kPa] 6.25 12.5 25 50 100 200 400 800 0.9 0.8 0.7 0.6 So we plot our data points. 0.5 0.4 1 10 100 1,000 Load [kPa]

Find: CC e Lab Test Data e Load [kPa] 0.9 0.8 0.7 0.6 0.5 0.4 1 10 100 0.919 0.900 0.890 0.868 0.825 0.771 0.687 0.572 0.463 Load [kPa] 6.25 12.5 25 50 100 200 400 800 0.9 0.8 0.7 0.6 --- 0.5 0.4 1 10 100 1,000 Load [kPa]

Find: CC e Lab Test Data e Load [kPa] 0.9 0.8 0.7 0.6 0.5 0.4 1 10 100 0.919 0.900 0.890 0.868 0.825 0.771 0.687 0.572 0.463 Load [kPa] 6.25 12.5 25 50 100 200 400 800 0.9 0.8 0.7 0.6 Keep in mind, this drawing is not to scale 0.5 0.4 1 10 100 1,000 Load [kPa]

Find: CC e Lab Test Data e Load [kPa] 0.9 0.8 0.7 0.6 0.5 0.4 1 10 100 0.919 0.900 0.890 0.868 0.825 0.771 0.687 0.572 0.463 Load [kPa] 6.25 12.5 25 50 100 200 400 800 0.9 0.8 0.7 0.6 --- 0.5 0.4 1 10 100 1,000 Load [kPa]

Find: CC e Lab Test Data e Load [kPa] 0.9 0.8 0.7 0.6 0.5 0.4 1 10 100 0.919 0.900 0.890 0.868 0.825 0.771 0.687 0.572 0.463 Load [kPa] 6.25 12.5 25 50 100 200 400 800 0.9 0.8 0.7 0.6 --- 0.5 0.4 1 10 100 1,000 Load [kPa]

Find: CC e Lab Test Data e Load [kPa] 0.9 0.8 0.7 0.6 0.5 0.4 1 10 100 0.919 0.900 0.890 0.868 0.825 0.771 0.687 0.572 0.463 Load [kPa] 6.25 12.5 25 50 100 200 400 800 0.9 0.8 0.7 0.6 --- 0.5 0.4 1 10 100 1,000 Load [kPa]

Find: CC e Lab Test Data e Load [kPa] 0.9 0.8 0.7 0.6 0.5 0.4 1 10 100 0.919 0.900 0.890 0.868 0.825 0.771 0.687 0.572 0.463 Load [kPa] 6.25 12.5 25 50 100 200 400 800 0.9 0.8 0.7 0.6 --- 0.5 0.4 1 10 100 1,000 Load [kPa]

Find: CC e Lab Test Data e Load [kPa] 0.9 0.8 0.7 0.6 0.5 0.4 1 10 100 0.919 0.900 0.890 0.868 0.825 0.771 0.687 0.572 0.463 Load [kPa] 6.25 12.5 25 50 100 200 400 800 0.9 0.8 0.7 0.6 --- 0.5 0.4 1 10 100 1,000 Load [kPa]

Find: CC e Lab Test Data e Load [kPa] 0.9 0.8 0.7 0.6 0.5 0.4 1 10 100 0.919 0.900 0.890 0.868 0.825 0.771 0.687 0.572 0.463 Load [kPa] 6.25 12.5 25 50 100 200 400 800 0.9 0.8 0.7 0.6 Once we’ve plotted all the points, we can compute the compression index, 0.5 0.4 1 10 100 1,000 Load [kPa]

Find: CC e Lab Test Data e Load [kPa] 0.9 0.8 0.7 0.6 0.5 0.4 1 10 100 0.919 0.900 0.890 0.868 0.825 0.771 0.687 0.572 0.463 Load [kPa] 6.25 12.5 25 50 100 200 400 800 0.9 0.8 0.7 0.6 which is the slope of the best fit line formed by the data points, at loading stresses greater than the preconsolidation stress. 0.5 0.4 1 10 100 1,000 Load [kPa]

σ’p Find: CC e Lab Test Data e Load [kPa] 0.9 0.8 0.7 0.6 0.5 0.4 1 10 0.919 0.900 0.890 0.868 0.825 0.771 0.687 0.572 0.463 Load [kPa] 6.25 12.5 25 50 100 200 400 800 0.9 0.8 0.7 0.6 So, if the preconsolidation stress is somewhere in here, 0.5 0.4 1 10 100 1,000 Load [kPa]

σ’p Find: CC e log 10 CC Lab Test Data e Load [kPa] 0.9 0.8 0.7 0.6 0.919 0.900 0.890 0.868 0.825 0.771 0.687 0.572 0.463 Load [kPa] 6.25 12.5 25 50 100 200 400 800 0.9 log 10 0.8 0.7 CC 0.6 we’re interested in the magnitude of the slope of the best fit line along these last three, or four, data points. Where log base 10 of 10, is equal to 1. 0.5 0.4 1 10 100 1,000 Load [kPa]

σ’p Find: CC e 1 CC Lab Test Data e Load [kPa] 0.9 0.8 0.7 0.6 0.5 0.4 0.919 0.900 0.890 0.868 0.825 0.771 0.687 0.572 0.463 Load [kPa] 6.25 12.5 25 50 100 200 400 800 0.9 1 0.8 0.7 CC 0.6 Suppose the best fit line passes most closely --- 0.5 0.4 1 10 100 1,000 Load [kPa]

σ’p Find: CC e 1 CC Lab Test Data e Load [kPa] 0.9 0.8 0.7 0.6 0.5 0.4 0.919 0.900 0.890 0.868 0.825 0.771 0.687 0.572 0.463 Load [kPa] 6.25 12.5 25 50 100 200 400 800 0.9 1 0.8 0.7 CC 0.6 through the two data points at loads of 200 and 400 kilopascals. 0.5 0.4 1 10 100 1,000 Load [kPa]

σ’p Find: CC e 1 CC e 0.9 0.8 0.7 Load [kPa] 0.6 0.5 0.4 1 10 100 0.687 0.572 Load [kPa] 200 400 0.6 We can compute the compression index, 0.5 0.4 1 10 100 1,000 Load [kPa]

σ’p Find: CC CC= e 1 CC e e400-e200 400 log 200 0.9 0.8 0.7 Load [kPa] 0.687 0.572 Load [kPa] 200 400 0.6 by computing the slope of the line through these two points. 0.5 0.4 1 10 100 1,000 Load [kPa]

σ’p Find: CC CC= e 1 CC=0.382 CC e e400-e200 400 log 200 0.9 0.8 0.7 0.687 0.572 Load [kPa] 200 400 0.6 The compression index is 0.382. [pause] 0.5 0.4 1 10 100 1,000 Load [kPa]

σ’p Find: CC CC= e 1 CC=0.382 CC e e400-e200 400 0.03 0.07 C) 0.38 log 200 0.9 1 0.8 CC=0.382 0.7 CC e 0.687 0.572 Load [kPa] 200 400 0.6 Returning to our possible solutions, 0.5 0.4 1 10 100 1,000 Load [kPa]

σ’p Find: CC CC= e 1 CC=0.382 AnswerC CC e e400-e200 400 0.03 0.07 log 200 0.9 1 0.8 CC=0.382 AnswerC 0.7 CC e 0.687 0.572 Load [kPa] 200 400 0.6 the answer is C. [pause] Now, if the problem happened to ask for the recompression index, 0.5 0.4 1 10 100 1,000 Load [kPa]

σ’p Find: CR e Lab Test Data e Load [kPa] 0.9 0.8 0.7 0.6 0.5 0.4 1 10 0.919 0.900 0.890 0.868 0.825 0.771 0.687 0.572 0.463 Load [kPa] 6.25 12.5 25 50 100 200 400 800 0.9 0.8 0.7 0.6 not the compression index, the same method can be used. 0.5 0.4 1 10 100 1,000 Load [kPa]

σ’p Find: CR e Lab Test Data e Load [kPa] 0.9 0.8 0.7 0.6 0.5 0.4 1 10 0.919 0.900 0.890 0.868 0.825 0.771 0.687 0.572 0.463 Load [kPa] 6.25 12.5 25 50 100 200 400 800 0.9 0.8 0.7 0.6 the only difference would be, we’d solve for the magnitude of the slope --- 0.5 0.4 1 10 100 1,000 Load [kPa]

σ’p Find: CR e 1 CR Lab Test Data e Load [kPa] 0.9 0.8 0.7 0.6 0.5 0.4 0.919 0.900 0.890 0.868 0.825 0.771 0.687 0.572 0.463 Load [kPa] 6.25 12.5 25 50 100 200 400 800 1 0.9 CR 0.8 0.7 0.6 of the best line passing through the data points with stress values less than the preconcolidation stress. 0.5 0.4 1 10 100 1,000 Load [kPa]

σ’p Find: CR e 1 CR Lab Test Data e Load [kPa] 0.9 0.8 0.7 0.6 0.5 0.4 0.919 0.900 0.890 0.868 Load [kPa] 6.25 12.5 25 1 0.9 CR 0.8 0.7 0.6 If we assume a line passing through the two data points at loads of 6.25 and 12.5 kilopascals is most parallel the best fit line, 0.5 0.4 1 10 100 1,000 Load [kPa]

σ’p Find: CR e 1 CR e12.5-e6.25 CR= Lab Test Data e 12.5 log 6.25 0.919 0.900 0.890 0.868 Load [kPa] 6.25 12.5 25 1 0.9 CR 0.8 0.7 e12.5-e6.25 CR= 0.6 we can plug the appropriate numbers into our equation, 12.5 log 0.5 6.25 0.4 1 10 100 1,000 Load [kPa]

σ’p Find: CR e 1 CR e12.5-e6.25 CR= Lab Test Data e 12.5 log 6.25 0.919 0.900 0.890 0.868 Load [kPa] 6.25 12.5 25 1 0.9 CR 0.8 0.7 e12.5-e6.25 CR= 0.6 And solve for a recompression index value of 0.033. 12.5 log 0.5 6.25 0.4 CR=0.033 1 10 100 1,000 Load [kPa]

( ) ? 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 ‘