Find: ρc [in] from load γT=106 [lb/ft3] γT=112 [lb/ft3]

Παρουσίαση με θέμα: "Find: ρc [in] from load γT=106 [lb/ft3] γT=112 [lb/ft3]"— Μεταγράφημα παρουσίασης:

1 Find: ρc [in] from load γT=106 [lb/ft3] γT=112 [lb/ft3]
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] γT=106 [lb/ft3] 4 [ft] 2.5 3.1 C) 3.7 D) 4.3 Sand γT=112 [lb/ft3] 2 [ft] γT=110 [lb/ft3] 20 [ft] Find the settlement from consolidation in inches from a load. In this problem, Clay CR=0.04 CC=0.40 γT=118 [lb/ft3] Gravel 4[ft]

2 Find: ρc [in] from load γT=106 [lb/ft3] γT=112 [lb/ft3]
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] γT=106 [lb/ft3] 4 [ft] 2.5 3.1 C) 3.7 D) 4.3 Sand γT=112 [lb/ft3] 2 [ft] γT=110 [lb/ft3] 20 [ft] a load of 400 pounds per square feet is applied to a soil profile, Clay CR=0.04 CC=0.40 γT=118 [lb/ft3] Gravel 4[ft]

3 Find: ρc [in] from load γT=106 [lb/ft3] γT=112 [lb/ft3]
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] γT=106 [lb/ft3] 4 [ft] 2.5 3.1 C) 3.7 D) 4.3 Sand γT=112 [lb/ft3] 2 [ft] γT=110 [lb/ft3] 20 [ft] consisting of 6 feet of sand, Clay CR=0.04 CC=0.40 γT=118 [lb/ft3] Gravel 4[ft]

4 Find: ρc [in] from load γT=106 [lb/ft3] γT=112 [lb/ft3]
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] γT=106 [lb/ft3] 4 [ft] 2.5 3.1 C) 3.7 D) 4.3 Sand γT=112 [lb/ft3] 2 [ft] γT=110 [lb/ft3] 20 [ft] 20 feet of clay, Clay CR=0.04 CC=0.40 γT=118 [lb/ft3] Gravel 4[ft]

5 Find: ρc [in] from load γT=106 [lb/ft3] γT=112 [lb/ft3]
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] γT=106 [lb/ft3] 4 [ft] 2.5 3.1 C) 3.7 D) 4.3 Sand γT=112 [lb/ft3] 2 [ft] γT=110 [lb/ft3] 20 [ft] and 4 feet of gravel. Clay CR=0.04 CC=0.40 γT=118 [lb/ft3] Gravel 4[ft]

6 Find: ρc [in] from load γT=106 [lb/ft3] γT=112 [lb/ft3]
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] γT=106 [lb/ft3] 4 [ft] 2.5 3.1 C) 3.7 D) 4.3 Sand γT=112 [lb/ft3] 2 [ft] γT=110 [lb/ft3] 20 [ft] We’ve also been given the preconsolidation stress for the entire soil profile. Clay CR=0.04 CC=0.40 γT=118 [lb/ft3] Gravel 4[ft]

7 Find: ρc [in] from load γT=106 [lb/ft3] γT=112 [lb/ft3]
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] γT=106 [lb/ft3] 4 [ft] 2.5 3.1 C) 3.7 D) 4.3 Sand γT=112 [lb/ft3] 2 [ft] γT=110 [lb/ft3] 20 [ft] Notice the groundwater table at a depth of 4 feet. Clay CR=0.04 CC=0.40 γT=118 [lb/ft3] Gravel 4[ft]

8 Find: ρc [in] from load γT=106 [lb/ft3] γT=112 [lb/ft3]
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] γT=106 [lb/ft3] 4 [ft] 2.5 3.1 C) 3.7 D) 4.3 Sand γT=112 [lb/ft3] 2 [ft] γT=110 [lb/ft3] 20 [ft] Consolidation will only occur in the clay layer because it is a fine grain soil. Clay CR=0.04 CC=0.40 γT=118 [lb/ft3] Gravel 4[ft]

9 Find: ρc [in] from load γT=106 [lb/ft3] γT=112 [lb/ft3]
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] 2 [ft] γT=110 [lb/ft3] Clay 20 [ft] --- CR=0.04 CC=0.40 γT=118 [lb/ft3] Grav 4[ft]

10 ( ) Find: ρc [in] from load H*C σfinal ρc= log 1+e σinitial
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] ( ) H*C σfinal ρc= 1+e σinitial log ‘ γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] 2 [ft] γT=110 [lb/ft3] Clay 20 [ft] Looking at our general consolidation settlement equation, CR=0.04 CC=0.40 γT=118 [lb/ft3] Grav 4[ft]

11 ( ) Find: ρc [in] from load H*C σfinal ρc= log 1+e σinitial
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] 20 [ft] ( ) H*C σfinal ρc= 1+e σinitial log ‘ γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] 2 [ft] γT=110 [lb/ft3] Clay 20 [ft] we know the height of the clay layer is 20 feet, CR=0.04 CC=0.40 γT=118 [lb/ft3] Grav 4[ft]

12 ( ) ? Find: ρc [in] from load H*C σfinal ρc= log 1+e σinitial
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] 20 [ft] ? ( ) H*C σfinal ρc= 1+e σinitial log ‘ γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] 2 [ft] γT=110 [lb/ft3] Clay 20 [ft] but we don’t know which compression index to use, CR=0.04 CC=0.40 γT=118 [lb/ft3] Grav 4[ft]

13 ( ) ? ? Find: ρc [in] from load H*C σfinal ρc= log 1+e σinitial
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] 20 [ft] ? ( ) H*C σfinal ρc= 1+e σinitial log ‘ γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] 2 [ft] ? γT=110 [lb/ft3] Clay 20 [ft] nor do we know the initial void ratio, CR=0.04 CC=0.40 γT=118 [lb/ft3] Grav 4[ft]

14 ( ) ? ? ? Find: ρc [in] from load H*C σfinal ρc= log 1+e σinitial
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] 20 [ft] ? ( ) H*C σfinal ρc= 1+e σinitial log ‘ γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] 2 [ft] ? ? γT=110 [lb/ft3] Clay 20 [ft] nor do we know the initial and final vertical effective stresses. CR=0.04 CC=0.40 γT=118 [lb/ft3] Grav 4[ft]

15 ( ) ? Find: ρc [in] from load H*C σfinal ρc= log 1+e σinitial
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] ( ) H*C σfinal ρc= 1+e σinitial log ‘ γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] ? 2 [ft] γT=110 [lb/ft3] Clay 20 [ft] Let’s first solve for the initial void ratio. CR=0.04 CC=0.40 γT=118 [lb/ft3] Grav 4[ft]

16 ? Find: ρc [in] from load γT= e γT=106 [lb/ft3] γT=112 [lb/ft3]
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] e ? (SG+e) * γW γT=106 [lb/ft3] 4 [ft] γT= Sand γT=112 [lb/ft3] 1+e 2 [ft] γT=110 [lb/ft3] Clay 20 [ft] From our equations, we know total unit weight of a saturated soil can be determined from the specific gravity, void ratio and unit weight of water. CR=0.04 CC=0.40 γT=118 [lb/ft3] Grav 4[ft]

17 ? Find: ρc [in] from load γT - γW γT= e γT=106 [lb/ft3]
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] e ? (SG+e) * γW γT=106 [lb/ft3] 4 [ft] γT= Sand γT=112 [lb/ft3] 1+e 2 [ft] SG * γW - γT e= γT - γW γT=110 [lb/ft3] Clay 20 [ft] Solving for the void ratio --- CR=0.04 CC=0.40 γT=118 [lb/ft3] Grav 4[ft]

18 ? Find: ρc [in] from load γT - γW γT= e γT=106 [lb/ft3]
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] e ? (SG+e) * γW γT=106 [lb/ft3] 4 [ft] γT= Sand γT=112 [lb/ft3] 1+e 2 [ft] 2.70 SG * γW - γT e= γT - γW γT=110 [lb/ft3] Clay 20 [ft] and plugging in our known values, assuming the specific gravity of clay is 2.7, CR=0.04 CC=0.40 62.4 [lb/ft3] γT=118 [lb/ft3] Grav 4[ft]

19 ? Find: ρc [in] from load γT - γW γT= e γT=106 [lb/ft3]
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] e ? (SG+e) * γW γT=106 [lb/ft3] 4 [ft] γT= Sand γT=112 [lb/ft3] 1+e 2 [ft] 2.70 SG * γW - γT e= γT - γW γT=110 [lb/ft3] Clay 20 [ft] the initial void ratio computes to [pause] CR=0.04 CC=0.40 62.4 [lb/ft3] e=1.23 γT=118 [lb/ft3] Grav 4[ft]

20 ( ) ? ? ? Find: ρc [in] from load H*C σfinal ρc= log 1+e σinitial
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] 20 [ft] ? ( ) H*C σfinal ρc= 1+e σinitial log ‘ γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] 2 [ft] ? ? γT=110 [lb/ft3] Clay 20 [ft] Returning to our settlement equation, CR=0.04 CC=0.40 e=1.23 γT=118 [lb/ft3] Grav 4[ft]

21 ( ) ? ? Find: ρc [in] from load H*C σfinal ρc= log 1+e σinitial
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] 20 [ft] ? ( ) H*C σfinal ρc= 1+e σinitial log ‘ γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] 2 [ft] 1.23 ? γT=110 [lb/ft3] Clay 20 [ft] we can now plug in our void ratio. CR=0.04 CC=0.40 γT=118 [lb/ft3] Grav 4[ft]

22 ( ) ? ? Find: ρc [in] from load H*C σfinal ρc= log 1+e σinitial
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] 20 [ft] ? ( ) H*C σfinal ρc= 1+e σinitial log ‘ γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] 2 [ft] 1.23 ? γT=110 [lb/ft3] Clay 20 [ft] Next we’ll determine which compression index to use. CR=0.04 CC=0.40 γT=118 [lb/ft3] Grav 4[ft]

23 ( ) ? ? Find: ρc [in] from load H*C σfinal ρc= log 1+e σinitial
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] 20 [ft] ? ( ) H*C σfinal ρc= 1+e σinitial log ‘ γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] 2 [ft] 1.23 ? γT=110 [lb/ft3] Clay 20 [ft] To do so, let’s first evaluate the initial and final vertical effective stress values --- CR=0.04 CC=0.40 γT=118 [lb/ft3] Grav 4[ft]

24 ( ) ? ? Find: ρc [in] from load H*C σfinal ρc= log 1+e σinitial
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] 20 [ft] ? ( ) H*C σfinal ρc= 1+e σinitial log ‘ γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] 2 [ft] 1.23 ? γT=110 [lb/ft3] Clay 20 [ft] at the midpoint of the clay layer. CR=0.04 CC=0.40 γT=118 [lb/ft3] Grav 4[ft]

25 ( ) ? ? Find: ρc [in] from load H*C σfinal ρc= log 1+e σinitial
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] 20 [ft] ? ( ) H*C σfinal ρc= 1+e σinitial log ‘ γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] 2 [ft] 1.23 ? σ’i = Σ γ’ * d γT=110 [lb/ft3] Clay 20 [ft] We add up the stress contribution --- CR=0.04 CC=0.40 γT=118 [lb/ft3] Grav 4[ft]

26 ( ) ? ? Find: ρc [in] from load H*C σfinal ρc= log 1+e σinitial
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] 20 [ft] ? ( ) H*C σfinal ρc= 1+e σinitial log ‘ γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] 2 [ft] 1.23 ? Clay σ’i,clay= 4 [ft] * 106 [lb/ft3] γT=110 [lb/ft3] 20 [ft] from the unsaturated sand, + … CR=0.04 CC=0.40 γT=118 [lb/ft3] Grav 4[ft]

27 ( ) ? ? Find: ρc [in] from load H*C σfinal ρc= log 1+e σinitial
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] 20 [ft] ? ( ) H*C σfinal ρc= 1+e σinitial log ‘ γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] 2 [ft] 1.23 ? Clay σ’i,clay= 4 [ft] * 106 [lb/ft3] γT=110 [lb/ft3] 20 [ft] from the saturated sand, + 2 [ft] * ( ) [lb/ft3] CR=0.04 CC=0.40 + … γT=118 [lb/ft3] Grav 4[ft]

28 ( ) ? ? Find: ρc [in] from load H*C σfinal ρc= log 1+e σinitial
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] 20 [ft] ? ( ) H*C σfinal ρc= 1+e σinitial log ‘ γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] 2 [ft] 1.23 ? Clay σ’i,clay= 4 [ft] * 106 [lb/ft3] γT=110 [lb/ft3] 20 [ft] and from the top 10 feet of clay. + 2 [ft] * ( ) [lb/ft3] CR=0.04 CC=0.40 +(10 [ft]) * ( ) [lb/ft3] γT=118 [lb/ft3] Grav 4[ft]

29 ( ) ? ? Find: ρc [in] from load H*C σfinal ρc= log 1+e σinitial
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] 20 [ft] ? ( ) H*C σfinal ρc= 1+e σinitial log ‘ γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] 2 [ft] 1.23 ? Clay σ’i,clay= 4 [ft] * 106 [lb/ft3] γT=110 [lb/ft3] 20 [ft] The initial stress here is 999 pounds per square feet. + 2 [ft] * ( ) [lb/ft3] CR=0.04 CC=0.40 +(10 [ft]) * ( ) [lb/ft3] σ’i,clay= 999 [lb/ft2] γT=118 [lb/ft3] Grav 4[ft]

30 ( ) ? ? Find: ρc [in] from load H*C σfinal ρc= log 1+e σinitial
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] 20 [ft] ? ( ) H*C σfinal ρc= 1+e σinitial log ‘ γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] 2 [ft] 1.23 ? Clay σ’i,clay= 4 [ft] * 106 [lb/ft3] γT=110 [lb/ft3] 20 [ft] After adding the load of 400 pounds per square feet, + 2 [ft] * ( ) [lb/ft3] CR=0.04 CC=0.40 +(10 [ft]) * ( ) [lb/ft3] σ’i,clay= 999 [lb/ft2] γT=118 [lb/ft3] Grav 4[ft]

31 ( ) ? ? Find: ρc [in] from load H*C σfinal ρc= log 1+e σinitial
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] 20 [ft] ? ( ) H*C σfinal ρc= 1+e σinitial log ‘ γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] 2 [ft] 1.23 ? Clay σ’i,clay= 4 [ft] * 106 [lb/ft3] γT=110 [lb/ft3] 20 [ft] the final stress 1,399 pounds the square feet. + 2 [ft] * ( ) [lb/ft3] CR=0.04 CC=0.40 +(10 [ft]) * ( ) [lb/ft3] σ’i,clay= 999 [lb/ft2] γT=118 [lb/ft3] σ’f,clay= 1,399 [lb/ft2] Grav 4[ft]

32 ( ) ? ? Find: ρc [in] from load σ’i < σ’p < σ’f H*C σfinal ρc=
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] 20 [ft] ? ( ) H*C σfinal ρc= 1+e σinitial log ‘ γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] 2 [ft] 1.23 ? Clay σ’i < σ’p < σ’f γT=110 [lb/ft3] 20 [ft] The preconsolidation stress is greater than the initial stress but less than the final stress. CR=0.04 CC=0.40 σ’i,clay= 999 [lb/ft2] γT=118 [lb/ft3] σ’f,clay= 1,399 [lb/ft2] Grav 4[ft]

33 ( ) ? ? Find: ρc [in] from load σ’i < σ’p < σ’f H*C σfinal ρc=
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] 20 [ft] ? ( ) H*C σfinal ρc= 1+e σinitial log ‘ γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] 2 [ft] 1.23 ? Clay σ’i < σ’p < σ’f γT=110 [lb/ft3] 20 [ft] Which means we’ll use both consolidation indices. [pause] CR=0.04 CC=0.40 σ’i,clay= 999 [lb/ft2] γT=118 [lb/ft3] σ’f,clay= 1,399 [lb/ft2] Grav 4[ft]

34 ( ) ? ? Find: ρc [in] from load H*C σfinal ρc= log 1+e σinitial
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] 20 [ft] ? ( ) H*C σfinal ρc= 1+e σinitial log ‘ γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] 2 [ft] 1.23 ? σ’i = Σ γ’ * d γT=110 [lb/ft3] Clay 20 [ft] At this point it would be efficient to derive an equation for the initial vertical effective stress in the clay layer as a function of depth. CR=0.04 CC=0.40 γT=118 [lb/ft3] Grav 4[ft]

35 ( ) ? ? Find: ρc [in] from load H*C σfinal ρc= log 1+e σinitial
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] 20 [ft] ? ( ) H*C σfinal ρc= 1+e σinitial log ‘ γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] 2 [ft] 1.23 ? Clay σ’i,clay= 4 [ft] * 106 [lb/ft3] γT=110 [lb/ft3] 20 [ft] As before, we add up the contribution from --- + … CR=0.04 CC=0.40 γT=118 [lb/ft3] Grav 4[ft]

36 ( ) ? ? Find: ρc [in] from load H*C σfinal ρc= log 1+e σinitial
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] 20 [ft] ? ( ) H*C σfinal ρc= 1+e σinitial log ‘ γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] 2 [ft] 1.23 ? Clay σ’i,clay= 4 [ft] * 106 [lb/ft3] γT=110 [lb/ft3] 20 [ft] the 6 feet of sand --- + 2 [ft] * ( ) [lb/ft3] CR=0.04 CC=0.40 + … γT=118 [lb/ft3] Grav 4[ft]

37 ( ) ? ? Find: ρc [in] from load H*C σfinal ρc= log 1+e σinitial
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] 20 [ft] ? ( ) H*C σfinal ρc= 1+e σinitial log ‘ γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] 2 [ft] 1.23 ? Clay σ’i,clay= 4 [ft] * 106 [lb/ft3] γT=110 [lb/ft3] 20 [ft] and from the clay, + 2 [ft] * ( ) [lb/ft3] CR=0.04 CC=0.40 +(d-6 [ft]) * ( ) [lb/ft3] γT=118 [lb/ft3] Grav 4[ft]

38 ( ) ? ? Find: ρc [in] from load H*C σfinal ρc= log 1+e σinitial
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] 20 [ft] ? ( ) H*C σfinal ρc= 1+e σinitial log ‘ γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] 2 [ft] 1.23 d=6[ft] ? Clay σ’i,clay= 4 [ft] * 106 [lb/ft3] γT=110 [lb/ft3] 20 [ft] Where we limit the depth ‘d’ between 6 feet to 26 feet. + 2 [ft] * ( ) [lb/ft3] CR=0.04 CC=0.40 +(d-6 [ft]) * ( ) [lb/ft3] d=26[ft] γT=118 [lb/ft3] Grav 4[ft]

39 ( ) ? ? Find: ρc [in] from load σ’i,clay= 238 [lb/ft2] H*C σfinal ρc=
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] 20 [ft] ? ( ) H*C σfinal ρc= 1+e σinitial log ‘ γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] 2 [ft] 1.23 d=6[ft] ? Clay σ’i,clay= 4 [ft] * 106 [lb/ft3] γT=110 [lb/ft3] 20 [ft] After solving, the initial stress in the clay layer is 238 pounds per square feet, plus ‘d’ times 47.6 pounds per cubic foot. + 2 [ft] * ( ) [lb/ft3] CR=0.04 CC=0.40 +(d-6 [ft]) * ( ) [lb/ft3] σ’i,clay= 238 [lb/ft2] d=26[ft] γT=118 [lb/ft3] +d * 47.6 [lb/ft3] Grav 4[ft]

40 ( ) ? ? Find: ρc [in] from load H*C σfinal ρc= log 1+e σinitial
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] 20 [ft] ? ( ) H*C σfinal ρc= 1+e σinitial log ‘ γT=106 [lb/ft3] 4 [ft] Sand γT=112 [lb/ft3] 2 [ft] 1.23 ? γT=110 [lb/ft3] Clay 20 [ft] Now, let’s zoom in on just the clay layer. CR=0.04 CC=0.40 γT=118 [lb/ft3] Grav 4[ft]

41 ( ) ? ? Find: ρc [in] from load H*C σfinal ρc= log 1+e σinitial
σ’p=1,200 [lb/ft2] 20 [ft] ? ( ) H*C σfinal ρc= 1+e σinitial log ‘ 1.23 ? γT=110 [lb/ft3] Clay 20 [ft] --- CR=0.04 CC=0.40

42 Find: ρc [in] from load Sand d=6[ft] Clay d=26[ft] Gravel
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] Sand d=6[ft] Clay 20 [ft] We’ll divide the clay layer into the segments which consolidate along --- d=26[ft] Gravel

43 R Find: ρc [in] from load Sand d=6[ft] d=26[ft] Gravel
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] Sand d=6[ft] R the recompression curve only, d=26[ft] Gravel

44 R C Find: ρc [in] from load Sand d=6[ft] d=26[ft] Gravel
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] Sand d=6[ft] R the virgin compression curve only, C d=26[ft] Gravel

45 + R R C C Find: ρc [in] from load Sand d=6[ft] d=26[ft] Gravel
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] Sand d=6[ft] R + R C and along both curves. [pause] C d=26[ft] Gravel

46 + R R C C Find: ρc [in] from load σ’f < σ’p Sand d=6[ft] d=26[ft]
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] Sand d=6[ft] σ’f < σ’p R + R C The soil that only consolidates along the recompression curve has a final stress less than the preconsolidated stress. C d=26[ft] Gravel

47 + R R C C Find: ρc [in] from load σ’f < σ’p σ’i > σ’p Sand
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] Sand d=6[ft] σ’f < σ’p R + R C And the soil that only consolidates along the virgin compression curve has an initial stress greater than the preconsolidated stress. C σ’i > σ’p d=26[ft] Gravel

48 + R R C C Find: ρc [in] from load σ’f < σ’p σ’i < σ’p < σ’p
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] Sand d=6[ft] σ’f < σ’p R σ’i < σ’p < σ’p + R C The remaining soil consolidates along both curves. σ’i > σ’p C d=26[ft] Gravel

49 + R R C C Find: ρc [in] from load σ’f < σ’p σ’i < σ’p < σ’p
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] Sand d=6[ft] σ’f < σ’p R d=? σ’i < σ’p < σ’p + R C The depth at which virgin compression begins, σ’i > σ’p C d=26[ft] Gravel

50 + R R C C Find: ρc [in] from load σ’f < σ’p σ’f=σ’p
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] Sand d=6[ft] σ’f < σ’p R σ’f=σ’p d=? σ’i < σ’p < σ’p + R C occurs where the final stress equals the preconsolidation stress. σ’i > σ’p C d=26[ft] Gravel

51 + R R C C Find: ρc [in] from load σ’f = σ’p σ’f < σ’p
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] Sand σ’f = σ’p d=6[ft] R σ’f < σ’p d=? + R C σ’i < σ’p < σ’p To determine this depth, we set the final stress equal to the preconsolidation stress, C σ’i > σ’p d=26[ft] Gravel

52 + R R C C Find: ρc [in] from load σ’f = σ’p σ’i + Load = σ’p
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] Sand σ’f = σ’p d=6[ft] R σ’i + Load = σ’p σ’f < σ’p d=? + R C σ’i < σ’p < σ’p and solve. The final stress equals the initial stress plus the load. C σ’i > σ’p d=26[ft] Gravel

53 + R R C C Find: ρc [in] from load σ’f = σ’p σ’i + Load = σ’p
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] Sand σ’f = σ’p d=6[ft] R σ’i + Load = σ’p σ’f < σ’p σ’i = σ’p - Load d=? + R C σ’i < σ’p < σ’p --- C σ’i > σ’p d=26[ft] Gravel

54 + R R C C Find: ρc [in] from load σ’f = σ’p σ’i + Load = σ’p
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] Sand σ’f = σ’p d=6[ft] R σ’i + Load = σ’p σ’f < σ’p σ’i = σ’p - Load d=? + R C σ’i= 238 [lb/ft2] σ’i < σ’p < σ’p We can use our previously derived equation for the initial stress as a function of depth, and, +d * 47.6 [lb/ft3] C σ’i > σ’p d=26[ft] Gravel

55 + R R C C Find: ρc [in] from load σ’f = σ’p σ’i + Load = σ’p
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] Sand σ’f = σ’p d=6[ft] R σ’i + Load = σ’p σ’f < σ’p σ’i = σ’p - Load d=? + R C σ’i= 238 [lb/ft2] σ’i < σ’p < σ’p after plugging in our known values, +d * 47.6 [lb/ft3] C σ’i > σ’p d=26[ft] Gravel

56 + R R C C Find: ρc [in] from load σ’f = σ’p σ’i + Load = σ’p
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] Sand σ’f = σ’p d=6[ft] R σ’i + Load = σ’p σ’f < σ’p σ’i = σ’p - Load d=? + R C σ’i= 238 [lb/ft2] σ’i < σ’p < σ’p we find ‘d’ equals feet. +d * 47.6 [lb/ft3] C d=11.81 [ft] σ’i > σ’p d=26[ft] Gravel

57 + R R C C Find: ρc [in] from load σ’f < σ’p σ’i < σ’p < σ’p
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] Sand 6 [ft] R σ’f < σ’p 11.81 [ft] + R C σ’i < σ’p < σ’p Next, the depth at which reconsolidation no longer occurs --- d=? C σ’i > σ’p 26 [ft] Gravel

58 + R R C C Find: ρc [in] from load σ’f < σ’p σ’i < σ’p < σ’p
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] Sand d=6[ft] R σ’f < σ’p 11.81 [ft] + R C σ’i < σ’p < σ’p is where the initial stress is equal to the preconsolidation stress. σ’i=σ’p d=? C σ’i > σ’p d=26[ft] Gravel

59 + R R C C Find: ρc [in] from load σ’i=σ’p σ’f < σ’p
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] Sand σ’i=σ’p d=6[ft] R σ’f < σ’p 11.81 [ft] + R C σ’i < σ’p < σ’p --- d=? C σ’i > σ’p d=26[ft] Gravel

60 + R R C C Find: ρc [in] from load σ’i=σ’p σ’f < σ’p
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] Sand σ’i=σ’p d=6[ft] R σ’i= 238 [lb/ft2] σ’f < σ’p 11.81 [ft] +d * 47.6 [lb/ft3] + R C σ’i < σ’p < σ’p Again, we use our equation relating initial stress to depth, d=? C σ’i > σ’p d=26[ft] Gravel

61 + R R C C Find: ρc [in] from load σ’i=σ’p σ’f < σ’p
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] Sand σ’i=σ’p d=6[ft] R σ’i= 238 [lb/ft2] σ’f < σ’p 11.81 [ft] +d * 47.6 [lb/ft3] + R C σ’i < σ’p < σ’p and substitute, and solve for ‘d.’ d=? C σ’i > σ’p d=26[ft] Gravel

62 + R R C C Find: ρc [in] from load σ’i=σ’p σ’f < σ’p
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] Sand σ’i=σ’p d=6[ft] R σ’i= 238 [lb/ft2] σ’f < σ’p 11.81 [ft] +d * 47.6 [lb/ft3] + R C σ’i < σ’p < σ’p d = [ft] Here, ‘d’ equals feet. d=? C σ’i > σ’p d=26[ft] Gravel

63 + R R C C Find: ρc [in] from load σ’i=σ’p σ’i= 238 [lb/ft2] Sand
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] Sand σ’i=σ’p d=6[ft] R σ’i= 238 [lb/ft2] 11.81 [ft] +d * 47.6 [lb/ft3] + R C d = [ft] --- 20.21[ft] C d=26[ft] Gravel

64 ( ) + ? R ? R C C Find: ρc [in] from load H*C σfinal ρc= log
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] 20 [ft] Sand ? d=6[ft] ( ) H*C σfinal ρc= 1+e σinitial log ‘ R 11.81 [ft] 1.23 ? + R C Returning to our consolidation settlement equation, 20.21[ft] C d=26[ft] Gravel

65 ( ) + ? R ? R C C Find: ρc [in] from load H*C σfinal ρc= log
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] 20 [ft] Sand ? d=6[ft] ( ) H*C σfinal ρc= 1+e σinitial log ‘ R 11.81 [ft] 1.23 ? + R C We realize we’ll have to evaluate the settlement in 3 parts, 20.21[ft] C d=26[ft] Gravel

66 ( ) + ? R ? R C C Find: ρc [in] from load H*C σfinal ρc= log
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] 20 [ft] Sand ? d=6[ft] ( ) H*C σfinal ρc= 1+e σinitial log ‘ R 11.81 [ft] 1.23 ? + R C and cannot use 20 feet as our value for H. 20.21[ft] C d=26[ft] Gravel

67 ( ) + ? R ? R C C Find: ρc [in] from load H*C σfinal ρc= log
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] Sand ? d=6[ft] ( ) H*C σfinal ρc= 1+e σinitial log ‘ H = 5.81 [ft] R 11.81 [ft] 1.23 ? H = 8.40 [ft] + R C We compute the heights of each part separately. 20.21[ft] H = 5.79 [ft] C d=26[ft] Gravel

68 ( ) + ? R ? R C C Find: ρc [in] from load H*C σfinal ρc= log
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] Sand ? d=6[ft] ( ) H*C σfinal ρc= 1+e σinitial log ‘ H = 5.81 [ft] R CR=0.04 11.81 [ft] 1.23 ? H = 8.40 [ft] + R C Also, we recall the consolidation indices, 20.21[ft] H = 5.79 [ft] C CC=0.40 d=26[ft] Gravel

69 ( ) + R ? R C C Find: ρc [in] from load H*C σfinal ρc= log
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] Sand d=6[ft] ( ) H*C σfinal ρc= 1+e σinitial log ‘ H = 5.81 [ft] R CR=0.04 11.81 [ft] 1.23 ? H = 8.40 [ft] + R C CR+C=0.22 and use the average consolidation index value for the soil which consolidates along both curves. 20.21[ft] H = 5.79 [ft] C CC=0.40 d=26[ft] Gravel

70 ( ) + R ? R C C Find: ρc [in] from load H*C σfinal ρc= log
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] Sand d=6[ft] ( ) H*C σfinal ρc= 1+e σinitial log ‘ H = 5.81 [ft] R CR=0.04 11.81 [ft] 1.23 ? H = 8.40 [ft] + R C CR+C=0.22 Lastly, we turn our attention to the initial and final effective stress values. 20.21[ft] H = 5.79 [ft] C CC=0.40 d=26[ft] Gravel

71 + R R C C Find: ρc [in] from load Sand d=6[ft] 11.81 [ft] 20.21[ft]
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] Sand d=6[ft] H = 5.81 [ft] R CR=0.04 11.81 [ft] H = 8.40 [ft] + R C CR+C=0.22 We’ll evaluate a point at the middle of each sub-layer, 20.21[ft] H = 5.79 [ft] C CC=0.40 d=26[ft] Gravel

72 + R R C C Find: ρc [in] from load Sand d=6[ft] 11.81 [ft] 20.21[ft]
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] Sand d=6[ft] H = 5.81 [ft] R dR=8.91 [ft] CR=0.04 11.81 [ft] H = 8.40 [ft] + R C dR+C=16.01 [ft] CR+C=0.22 determine the depth of each point, 20.21[ft] H = 5.79 [ft] C dC=23.11 [ft] CC=0.40 d=26[ft] Gravel

73 + R R C C Find: ρc [in] from load Sand d=6[ft] 11.81 [ft] 20.21[ft]
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] Sand dR=8.91 [ft] d=6[ft] H = 5.81 [ft] dR+C=16.01 [ft] R CR=0.04 dC=23.11 [ft] 11.81 [ft] H = 8.40 [ft] + R C CR+C=0.22 --- 20.21[ft] H = 5.79 [ft] C CC=0.40 d=26[ft] Gravel

74 + R R C C Find: ρc [in] from load σ’i= 238 [lb/ft2] σ’f= σ’i + Load
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] Sand dR=8.91 [ft] d=6[ft] H = 5.81 [ft] dR+C=16.01 [ft] R CR=0.04 dC=23.11 [ft] 11.81 [ft] σ’i= 238 [lb/ft2] H = 8.40 [ft] + R C CR+C=0.22 +d * 47.6 [lb/ft3] and compute the initial and final stress values, 20.21[ft] σ’f= σ’i + Load H = 5.79 [ft] C CC=0.40 d=26[ft] Gravel

75 + R R C C Find: ρc [in] from load σ’i= 238 [lb/ft2] σ’f= σ’i + Load
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] Sand dR=8.91 [ft] d=6[ft] H = 5.81 [ft] dR+C=16.01 [ft] R CR=0.04 dC=23.11 [ft] 11.81 [ft] σ’i= 238 [lb/ft2] H = 8.40 [ft] + R C CR+C=0.22 +d * 47.6 [lb/ft3] since we know the depths and we know the load. 20.21[ft] σ’f= σ’i + Load H = 5.79 [ft] C CC=0.40 d=26[ft] Gravel

76 + R R C C Find: ρc [in] from load Sand d=6[ft] 11.81 [ft] 20.21[ft]
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] Sand d=6[ft] σ’i = [lb/ft2] H = 5.81 [ft] R CR=0.04 σ’f = 1,062.1 [lb/ft2] 11.81 [ft] σ’i = 1,400.1 [lb/ft2] H = 8.40 [ft] + R C σ’f = 1,800.1 [lb/ft2] CR+C=0.22 [pause] Finally, we compute the settlement for each layer. [pause] 20.21[ft] σ’i = 1,338.0 [lb/ft2] H = 5.79 [ft] C σ’f = 1,738.0 [lb/ft2] CC=0.40 d=26[ft] Gravel

77 ( ) R Find: ρc [in] from load H*C σfinal ρc= log 1+e σinitial e=1.23
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] Sand d=6[ft] σ’i = [lb/ft2] H = 5.81 [ft] R CR=0.04 σ’f = 1,062.1 [lb/ft2] 11.81 [ft] e=1.23 ( ) H*C σfinal ρc= 1+e σinitial log ‘ [pause] For the top layer,

78 ( ) R Find: ρc [in] from load H*C σfinal ρc= log 1+e σinitial e=1.23
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] Sand d=6[ft] σ’i = [lb/ft2] H = 5.81 [ft] R CR=0.04 σ’f = 1,062.1 [lb/ft2] 11.81 [ft] e=1.23 ( ) H*C σfinal ρc= 1+e σinitial log ‘ we compute 0.26 inches. [pause] ρc=0.26 [in]

79 ( ) + R C Find: ρc [in] from load H*C σfinal ρc= log 1+e σinitial
σ’p=1,200 [lb/ft2] Load=400 [lb/ft2] ( ) H*C σfinal ρc= 1+e σinitial log ‘ =1.45 [in] e=1.23 11.81 [ft] σ’i = 1,400.1 [lb/ft2] H = 8.40 [ft] + R C σ’f = 1,800.1 [lb/ft2] CR+C=0.22 This process is repeated for the middle layer, 20.21[ft]

80 ( ) C Find: ρc [in] from load H*C σfinal ρc= log 1+e σinitial
σ’p=1,200 [lb/ft2] Load=400 [lb/ft2] ( ) H*C σfinal ρc= 1+e σinitial log ‘ =1.42 [in] e=1.23 and for the bottom layer. 20.21[ft] σ’i = 1,338.0 [lb/ft2] H = 5.79 [ft] C σ’f = 1,738.0 [lb/ft2] CC=0.40 d=26[ft] Gravel

81 + R R C C Find: ρc [in] from load Sand d=6[ft] ρc=0.26 [in] 11.81 [ft]
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] Sand d=6[ft] H = 5.81 [ft] R ρc=0.26 [in] CR=0.04 11.81 [ft] H = 8.40 [ft] + ρc=1.45 [in] R C CR+C=0.22 The total consolidation from the load is, 20.21[ft] H = 5.79 [ft] C ρc=1.42 [in] CC=0.40 d=26[ft] Gravel

82 + R R C C Find: ρc [in] from load Sand d=6[ft] ρc=0.26 [in] 11.81 [ft]
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] Sand d=6[ft] H = 5.81 [ft] R ρc=0.26 [in] CR=0.04 11.81 [ft] H = 8.40 [ft] + ρc=1.45 [in] R C CR+C=0.22 3.13 inches. 20.21[ft] H = 5.79 [ft] C ρc=1.42 [in] CC=0.40 d=26[ft] ρc=3.13 [in] Gravel

83 + R R C C Find: ρc [in] from load Sand d=6[ft] ρc=0.26 [in] 2.5 3.1
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] Sand d=6[ft] H = 5.81 [ft] R ρc=0.26 [in] CR=0.04 2.5 3.1 C) 3.7 D) 4.3 11.81 [ft] H = 8.40 [ft] + ρc=1.45 [in] R C CR+C=0.22 Returning to our possible solutions, 20.21[ft] H = 5.79 [ft] C ρc=1.42 [in] CC=0.40 d=26[ft] ρc=3.13 [in] Gravel

84 + R R C C Find: ρc [in] from load Sand d=6[ft] ρc=0.26 [in] 2.5 3.1
Load=400 [lb/ft2] σ’p=1,200 [lb/ft2] Sand d=6[ft] H = 5.81 [ft] R ρc=0.26 [in] CR=0.04 2.5 3.1 C) 3.7 D) 4.3 11.81 [ft] H = 8.40 [ft] + ρc=1.45 [in] R C CR+C=0.22 the answer is B. 20.21[ft] H = 5.79 [ft] C ρc=1.42 [in] CC=0.40 d=26[ft] ρc=3.13 [in] Gravel Answer  B

85 ( ) ? 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 ‘