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Find: the soil classification
SL=5 PL=30 LL=35 gravel=10% sand=35% fines=55% inorganic PI LL 100 60 40 20 80 10 30 50 CL CH ML MH ML OL CH SW Find the soil classification. [pause] In this problem, ---
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Find: the soil classification
SL=5 PL=30 LL=35 gravel=10% sand=35% fines=55% inorganic PI LL 100 60 40 20 80 10 30 50 CL CH ML MH ML OL CH SW we’ve been provided the Atterberg limits, the percentage of each soil type by weight, and informed the soil contains no organic material.
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Find: the soil classification
SL=5 PL=30 LL=35 gravel=10% sand=35% fines=55% inorganic PI LL 100 60 40 20 80 10 30 50 CL CH ML MH ML OL CH SW To determine whether or not this soil ---
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Find: the soil classification
SL=5 PL=30 LL=35 gravel=10% sand=35% fines=55% inorganic PI LL 100 60 40 20 80 10 30 50 CL CH ML MH ML OL CH SW is a fine grain soil or a coarse grain soil, ---
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Find: the soil classification
SL=5 PL=30 LL=35 gravel=10% sand=35% fines=55% inorganic PI LL 100 60 40 20 80 10 30 50 CL CH ML MH ML OL CH SW we add the percentages of gravel and sand ---
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Find: the soil classification
SL=5 PL=30 LL=35 gravel=10% sand=35% fines=55% inorganic 45% PI LL 100 60 40 20 80 10 30 50 CL CH ML MH ML OL CH SW and get 45%, Since there is more fine grain soil ---
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Find: the soil classification
SL=5 PL=30 LL=35 gravel=10% sand=35% fines=55% inorganic 45% PI LL 100 60 40 20 80 10 30 50 CL CH ML MH ML OL CH SW that coarse grain soil, by weight, we know we’re dealing with ---
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Find: the soil classification
SL=5 PL=30 LL=35 gravel=10% sand=35% fines=55% inorganic 45% fine grain soil PI LL 100 60 40 20 80 10 30 50 CL CH ML MH ML OL CH SW a fine gain soil, either a silt or a clay, and
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Find: the soil classification
SL=5 PL=30 LL=35 gravel=10% sand=35% fines=55% inorganic 45% fine grain soil PI LL 100 60 40 20 80 10 30 50 CL CH ML MH ML OL CH SW we know the soil is inorganic. [pause] Next, we look at the Atterberg limits ---
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Find: the soil classification
SL=5 PL=30 LL=35 gravel=10% sand=35% fines=55% inorganic 45% fine grain soil PI LL 100 60 40 20 80 10 30 50 CL CH ML MH ML OL CH SW and determine the plasticity index value, which is equal to ---
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Find: the soil classification
SL=5 PL=30 LL=35 gravel=10% sand=35% fines=55% inorganic 45% fine grain soil PI LL 100 60 40 20 80 10 30 50 PI=LL-PL ML OL CH SW CL CH ML MH the liquid limit minus the plastic limit.
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Find: the soil classification
SL=5 PL=30 LL=35 gravel=10% sand=35% fines=55% inorganic 45% fine grain soil PI LL 100 60 40 20 80 10 30 50 PI=LL-PL ML OL CH SW CL CH ML MH or 35, ---
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Find: the soil classification
SL=5 PL=30 LL=35 gravel=10% sand=35% fines=55% inorganic 45% fine grain soil PI LL 100 60 40 20 80 10 30 50 PI=LL-PL ML OL CH SW CL CH ML MH minus 30.
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Find: the soil classification
SL=5 PL=30 LL=35 gravel=10% sand=35% fines=55% inorganic 45% fine grain soil PI LL 100 60 40 20 80 10 30 50 PI=LL-PL PI=5 ML OL CH SW CL CH ML MH which gives us 5.
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Find: the soil classification
SL=5 PL=30 LL=35 gravel=10% sand=35% fines=55% inorganic 45% fine grain soil PI LL 100 60 40 20 80 10 30 50 PI=LL-PL PI=5 ML OL CH SW CL CH ML MH The point at 35, 5 on the plasticity chart falls in the ML region, signifying the soil is a low plasticity silt.
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Find: the soil classification
SL=5 PL=30 LL=35 gravel=10% sand=35% fines=55% inorganic 45% fine grain soil PI LL 100 60 40 20 80 10 30 50 PI=LL-PL PI=5 ML OL CH SW CL CH ML MH Looking at our possible solutions,
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Find: the soil classification
SL=5 PL=30 LL=35 gravel=10% sand=35% fines=55% inorganic 45% fine grain soil PI LL 100 60 40 20 80 10 30 50 PI=LL-PL PI=5 ML OL CH SW CL CH ML MH the answer is A AnswerA
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( ) ? τ [lb/ft2] γclay=53.1[lb/ft3] Index σ’v = Σ φ γ Δ d ˚ H*C σfinal
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] (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
![( ) τ [lb/ft2] γclay=53.1[lb/ft3] Index σ’v = Σ φ γ Δ d ˚ H*C σfinal](http://slideplayer.com./slide/15748395/88/images/19/%28+%29+%CF%84+%5Blb%2Fft2%5D+%CE%B3clay%3D53.1%5Blb%2Fft3%5D+Index+%CF%83%E2%80%99v+%3D+%CE%A3+%CF%86+%CE%B3+%CE%94+d+%CB%9A+H%2AC+%CF%83final.jpg "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] (5 [cm])2 * π/4. ( ) H*C σfinal. ρcn= 1+e σinitial. log. ‘ φ. size[mm] % passing. 0% 20% 40% 60% 80% 100% c= ,400. σ3. Sand. σ1.")
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