Find: Saturation, S 11% d=2.8 [in] 17% 23% 83% L=5.5 [in] clay

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Find: Saturation, S 11% d=2.8 [in] 17% 23% 83% L=5.5 [in] clay Wtotal=2.58 [lb] SG=2.75 wc=15% Find the saturation of the sample of clay. In this problem,

Find: Saturation, S 11% d=2.8 [in] 17% 23% 83% L=5.5 [in] clay Wtotal=2.58 [lb] SG=2.75 wc=15% a cylindrical sample of clay is provided with it’s known dimensions,

Find: Saturation, S 11% d=2.8 [in] 17% 23% 83% L=5.5 [in] clay Wtotal=2.58 [lb] SG=2.75 wc=15% as well as it’s total weight, specific gravity and water content.

Find: Saturation, S VW S= VV d=2.8 [in] L=5.5 [in] clay Wtotal=2.58 [lb] SG=2.75 wc=15% Saturation is defined as the volume of water divided by the volume of voids. To solve for these two variables,

Find: Saturation, S clay A W S Wtotal=2.58 [lb] SG=2.75 wc=15% V [ft3] W [lb] we’ll need to set up a block block diagram for the sample of clay. clay Wtotal=2.58 [lb] SG=2.75 wc=15% L=5.5 [in] d=2.8 [in]

Find: Saturation, S clay A W S 2.58 Wtotal=2.58 [lb] SG=2.75 wc=15% V [ft3] W [lb] 2.58 We were given the total weight is 2.58 pounds. clay Wtotal=2.58 [lb] SG=2.75 wc=15% L=5.5 [in] d=2.8 [in]

Find: Saturation, S clay π * d2 VT = L * A 4 W S 2.58 Wtotal=2.58 [lb] V [ft3] W [lb] π * d2 VT = L * 4 2.58 and, we can calculate the total volume by knowing, the cylinder’s length and diameter. clay Wtotal=2.58 [lb] SG=2.75 wc=15% L=5.5 [in] d=2.8 [in]

Find: Saturation, S clay π * d2 VT = L * A 4 W S VT = 0.0196 [ft3] V [ft3] W [lb] π * d2 VT = L * 4 VT = 0.0196 [ft3] 0.0196 2.58 The total volume of the sample is 0.0196 cubic feet. clay Wtotal=2.58 [lb] SG=2.75 wc=15% L=5.5 [in] d=2.8 [in]

Find: Saturation, S clay A W S 0.0196 2.58 Wtotal=2.58 [lb] V [ft3] W [lb] 0.0196 2.58 Since we know the weight of air is zero, and we know the total weight, clay Wtotal=2.58 [lb] SG=2.75 wc=15% L=5.5 [in] d=2.8 [in]

Find: Saturation, S clay wc= A W S WW WS 0.0196 2.58 Wtotal=2.58 [lb] V [ft3] W [lb] wc= WW WS WW WS 0.0196 2.58 we can use the water content to solve for the weight of water and weight of solid. clay Wtotal=2.58 [lb] SG=2.75 wc=15% L=5.5 [in] d=2.8 [in]

Find: Saturation, S clay wc= A W WW=wc * WS S WW WS 0.0196 2.58 V [ft3] W [lb] wc= WW WS WW=wc * WS WW WS 0.0196 2.58 The weight of water is 15% the weight of the solid. clay Wtotal=2.58 [lb] SG=2.75 wc=15% L=5.5 [in] d=2.8 [in]

Find: Saturation, S clay wc= A W WW=wc * WS S WW=0.15 * WS WW WS V [ft3] W [lb] wc= WW WS WW=wc * WS WW WW=0.15 * WS WS 0.0196 2.58 We also know --- clay Wtotal=2.58 [lb] SG=2.75 wc=15% L=5.5 [in] d=2.8 [in]

Find: Saturation, S clay wc= A W WW=wc * WS S WW=0.15 * WS V [ft3] W [lb] wc= WW WS WW=wc * WS WW WW=0.15 * WS WS WT = WW + WS 0.0196 2.58 the total weight equals the weight of water plus the weight of solid. clay Wtotal=2.58 [lb] SG=2.75 wc=15% L=5.5 [in] d=2.8 [in]

Find: Saturation, S clay wc= A W WW=wc * WS S WW=0.15 * WS V [ft3] W [lb] wc= WW WS WW=wc * WS WW WW=0.15 * WS WS WT = WW + WS 0.0196 2.58 WS = WT - WW Rearranging the equation, clay Wtotal=2.58 [lb] SG=2.75 wc=15% L=5.5 [in] d=2.8 [in]

Find: Saturation, S clay wc= A W WW=wc * WS S WW=0.15 * WS V [ft3] W [lb] wc= WW WS WW=wc * WS WW WW=0.15 * WS WS WT = WW + WS 0.0196 2.58 WS = WT - WW the weight of solid is 2.58 pounds minus the weight of water. clay Wtotal=2.58 [lb] SG=2.75 wc=15% L=5.5 [in] d=2.8 [in]

Find: Saturation, S clay wc= A W WW=wc * WS S WW=0.15 * WS V [ft3] W [lb] wc= WW WS WW=wc * WS WW WW=0.15 * WS WS WT = WW + WS 0.0196 2.58 WS = WT - WW If we plug this into a previous equation and skip a couple steps of algebra, WS = 2.58 [lb] - WW clay Wtotal=2.58 [lb] SG=2.75 wc=15% L=5.5 [in] d=2.8 [in]

Find: Saturation, S clay wc= A W WW=wc * WS S WW=0.15 * WS V [ft3] W [lb] wc= WW WS WW=wc * WS 0.337 WW=0.15 * WS WS WT = WW + WS 0.0196 2.58 WS = WT - WW we find the weight of water is 0.337 pounds. WS = 2.58 [lb] - WW clay Wtotal=2.58 [lb] SG=2.75 wc=15% 1.15 * WW = 0.387 [lb] WW = 0.337 [lb] L=5.5 [in] d=2.8 [in]

Find: Saturation, S clay wc= A W WW=wc * WS S WW=0.15 * WS V [ft3] W [lb] wc= WW WS WW=wc * WS 0.337 WW=0.15 * WS 2.243 WT = WW + WS 0.0196 2.58 WS = WT - WW And by quick subtraction, the weight of solids is 2.243 pounds. [pause] WS = 2.58 [lb] - WW clay Wtotal=2.58 [lb] SG=2.75 wc=15% 1.15 * WW = 0.387 [lb] WW = 0.337 [lb] WS = 2.243 [lb] L=5.5 [in] d=2.8 [in]

γS Find: Saturation, S clay A W S WS VS= VA VW 0.337 VS 2.243 0.0196 V [ft3] W [lb] WS VS= γS VA VW 0.337 VS 2.243 0.0196 2.58 Next, we solve for the volumes of air, water and solid. clay Wtotal=2.58 [lb] SG=2.75 wc=15% L=5.5 [in] d=2.8 [in]

γS γW Find: Saturation, S clay A W S WS 2.243 [lb] VS= = VA 62.4 VW V [ft3] W [lb] WS 2.243 [lb] VS= γS = VA lb 62.4 *2.75 ft3 γW VW 0.337 VS 2.243 0.0196 2.58 The volume of solid is the weight of solid divided the unit weigh of solid, clay Wtotal=2.58 [lb] SG=2.75 wc=15% L=5.5 [in] d=2.8 [in]

γS γW Find: Saturation, S clay A W S WS 2.243 [lb] VS= = VA 62.4 VW V [ft3] W [lb] WS 2.243 [lb] VS= γS = VA lb 62.4 *2.75 ft3 γW VW 0.337 VS=0.0131 [ft3] VS 2.243 0.0196 2.58 and equals 0.0131 cubic feet. clay Wtotal=2.58 [lb] SG=2.75 wc=15% L=5.5 [in] d=2.8 [in]

γS γW γW Find: Saturation, S clay A W S WS 2.243 [lb] VS= = VA 62.4 VW V [ft3] W [lb] WS 2.243 [lb] VS= γS = VA lb 62.4 *2.75 ft3 γW VW 0.337 VS=0.0131 [ft3] 0.0131 2.243 WW 0.337 [lb] VW= γW = 0.0196 2.58 lb 62.4 The volume of water is solved much the same, ft3 clay Wtotal=2.58 [lb] SG=2.75 wc=15% L=5.5 [in] d=2.8 [in]

γS γW γW Find: Saturation, S clay A W S WS 2.243 [lb] VS= = VA 62.4 VW V [ft3] W [lb] WS 2.243 [lb] VS= γS = VA lb 62.4 *2.75 ft3 γW VW 0.337 VS=0.0131 [ft3] 0.0131 2.243 WW 0.337 [lb] VW= γW = 0.0196 2.58 lb 62.4 and equals 0.0054 cubic feet. ft3 clay Wtotal=2.58 [lb] SG=2.75 wc=15% VW=0.0054 [ft3] L=5.5 [in] d=2.8 [in]

γS γW γW Find: Saturation, S clay A W S WS 2.243 [lb] VS= = VA 62.4 V [ft3] W [lb] WS 2.243 [lb] VS= γS = VA lb 62.4 *2.75 ft3 γW 0.0054 0.337 VS=0.0131 [ft3] 0.0131 2.243 WW 0.337 [lb] VW= γW = 0.0196 2.58 lb 62.4 The volume of air is the total volume minus the volume of solid minus the volume of water, ft3 clay Wtotal=2.58 [lb] SG=2.75 wc=15% VW=0.0054 [ft3] VA = VT - VS - VW L=5.5 [in] d=2.8 [in]

γS γW γW Find: Saturation, S clay A W S WS 2.243 [lb] VS= = VA 62.4 V [ft3] W [lb] WS 2.243 [lb] VS= γS = VA lb 62.4 *2.75 ft3 γW 0.0054 0.337 VS=0.0131 [ft3] 0.0131 2.243 WW 0.337 [lb] VW= γW = 0.0196 2.58 lb 62.4 and equals 0.0011 cubic feet. ft3 clay Wtotal=2.58 [lb] SG=2.75 wc=15% VW=0.0054 [ft3] VA = VT - VS - VW VA = 0.0011 [ft3] L=5.5 [in] d=2.8 [in]

Find: Saturation, S VW S= VV clay A W S 0.0011 0.0054 0.337 0.0131 V [ft3] W [lb] VW S= 0.0011 VV 0.0054 0.337 0.0131 2.243 0.0196 2.58 With our block diagram solved, we can compute the saturation of the soil. clay Wtotal=2.58 [lb] SG=2.75 wc=15% L=5.5 [in] d=2.8 [in]

Find:Saturation, S VW VW S= = VV VW+VA clay A W S 0.0011 0.0054 0.337 V [ft3] W [lb] VW VW S= = 0.0011 VV VW+VA 0.0054 0.337 0.0131 2.243 0.0196 2.58 After plugging in the appropriate values, clay Wtotal=2.58 [lb] SG=2.75 wc=15% L=5.5 [in] d=2.8 [in]

Find:Saturation, S VW VW S= = VV VW+VA S = 83.1% clay A W S 0.0011 V [ft3] W [lb] VW VW S= = 0.0011 VV VW+VA 0.0054 0.337 S = 83.1% 0.0131 2.243 0.0196 2.58 we find the saturation is 83.1 percent. clay Wtotal=2.58 [lb] SG=2.75 wc=15% L=5.5 [in] d=2.8 [in]

Find:Saturation, S VW VW S= = VV VW+VA S = 83.1% clay A W S 0.0011 V [ft3] W [lb] VW VW S= = 0.0011 VV VW+VA 0.0054 0.337 S = 83.1% 0.0131 2.243 0.0196 2.58 11% 17% 23% 83% When reviewing the possible solutions, clay Wtotal=2.58 [lb] SG=2.75 wc=15% L=5.5 [in] d=2.8 [in]

Find:Saturation, S VW VW S= = VV VW+VA S = 83.1% AnswerD clay A W S V [ft3] W [lb] VW VW S= = 0.0011 VV VW+VA 0.0054 0.337 S = 83.1% 0.0131 2.243 AnswerD 0.0196 2.58 11% 17% 23% 83% the answer is D. clay Wtotal=2.58 [lb] SG=2.75 wc=15% L=5.5 [in] d=2.8 [in]

( ) ? τ [lb/ft2] γclay=53.1[lb/ft3] Index σ’v = Σ φ γ Δ d ˚ H*C σfinal wc= WW WS 1 Index γclay=53.1[lb/ft3] Find: σ’v ρc d = 30 feet (1+wc)*γw wc+(1/SG) σ’v = Σ φ γ Δ d ˚ d Sand 10 ft γT=100 [lb/ft3] 100 [lb/ft3] φ=α1-α2 10 [ft] 20 ft Clay = γsand dsand +γclay dclay A W S V [ft3] W [lb] 40 ft text wc = 37% ? Δh 20 [ft] τ [lb/ft2] 62.4 lb ft3 27 yd3 ft3 (5 [cm])2 * π/4 ( ) H*C σfinal ρcn= 1+e σinitial log ‘ φ size[mm] % passing 0% 20% 40% 60% 80% 100% 10 1 0.1 0.01 c=0 400 1,400 σ3 Sand σ1