γdry=110 Find: VTotal[ft3] of borrowed soil Fill VT=10,000 [ft3]

γdry=110 Find: VTotal[ft3] of borrowed soil Fill VT=10,000 [ft3]
wc=10% Borrow wc=10% S=50% SG=2.67 γdry=110 lb ft3 A W S V [ft3] W [lb] 371 9,872 10,000 10,128 Find the total volume of borrowed soil, in cubic feet. [pause] In this problem, ---

wc=10% Borrow wc=10% S=50% SG=2.67 γdry=110 lb ft3 A W S V [ft3] W [lb] 371 9,872 10,000 10,128 soil is being transported from a borrow site to a fill site, and specific soil parameters ---

wc=10% Borrow wc=10% S=50% SG=2.67 γdry=110 lb ft3 A W S V [ft3] W [lb] 371 9,872 10,000 10,128 are provided. [pause] The problem asks to find the ---

γdry=110 ? Find: VTotal[ft3] of borrowed soil Fill VT=10,000 [ft3]
wc=10% Borrow wc=10% S=50% SG=2.67 γdry=110 lb ft3 A W S V [ft3] W [lb] 371 9,872 10,000 10,128 total volume of the borrowed material, which we’ll designate as V T. We’ll also abbreviate the other quantities ---- ? VT

wc=10% Borrow wc=10% S=50% SG=2.67 γdry=110 lb ft3 A W S V [ft3] W [lb] VA WA 371 9,872 10,000 10,128 of the block diagram as shown. [pause] In borrow-fill type problems, we usually have separate block diagrams --- VW WW VS WS VT WT

wc=10% Borrow wc=10% S=50% SG=2.67 γdry=110 lb ft3 A W S V [ft3] W [lb] A W S V [ft3] W [lb] VA WA VA WA for the borrowed material, and filled material, since the specified fill material rarely has the exact same ---- VW WW VW WW VS WS VS WS VT WT VT WT

γdry=110 = Find: VTotal[ft3] of borrowed soil Fill VT=10,000 [ft3]
wc=10% Borrow wc=10% S=50% SG=2.67 γdry=110 lb ft3 A W S V [ft3] W [lb] A W S V [ft3] W [lb] VA WA VA WA soil properties, as the nearby borrow pit. Also, for clarification, a subscript F or B, ---- VW WW VW WW VS WS VS WS VT WT VT WT

wc=10% Borrow wc=10% S=50% SG=2.67 γdry=110 lb ft3 A W S V [ft3] W [lb] A W S V [ft3] W [lb] VA,F WA,F VA,B WA,B will be added to the variable letters to identify this difference. In this problem, the volumes and weights of --- VW,F WW,F VW,B WW,B VS,F WS,F VS,B WS,B VT,F WT,F VT,B WT,B

wc=10% Borrow wc=10% S=50% SG=2.67 γdry=110 lb ft3 A W S V [ft3] W [lb] A W S V [ft3] W [lb] VA,F WA,F VA,B WA,B solid material is equal, therefore, therefore we can drop these sub-script terms, --- VW,F WW,F VW,B WW,B VS,F WS,F VS,B WS,B VT,F WT,F VT,B WT,B

wc=10% Borrow wc=10% S=50% SG=2.67 γdry=110 lb ft3 A W S V [ft3] W [lb] A W S V [ft3] W [lb] VA,F WA,F VA,B WA,B F and B. [pause] To find the total volume of borrowed material, --- VW,F WW,F VW,B WW,B VS WS VS WS VT,F WT,F VT,B WT,B

wc=10% Borrow wc=10% S=50% SG=2.67 γdry=110 lb ft3 A W S V [ft3] W [lb] A W S V [ft3] W [lb] VA,F WA,F VA,B WA,B we’ll start with our 10,000 cubic feet of total fill volume, ---- VW,F WW,F VW,B WW,B VS WS VS WS VT,F WT,F VT,B WT,B

wc=10% Borrow wc=10% S=50% SG=2.67 γdry=110 lb ft3 A W S V [ft3] W [lb] A W S V [ft3] W [lb] VA,F WA,F VA,B WA,B in the fill block-diagram. Next we’ll find the weight of the solid material, ---- VW,F WW,F VW,B WW,B VS WS VS WS WT,F VT,B WT,B 10,000

wc=10% Borrow wc=10% S=50% SG=2.67 γdry=110 lb ft3 A W S V [ft3] W [lb] A W S V [ft3] W [lb] VA,F WA,F VA,B WA,B which equals the weight of solid material in the --- VW,F WW,F VW,B WW,B VS WS VS WS WT,F VT,B WT,B 10,000

wc=10% Borrow wc=10% S=50% SG=2.67 γdry=110 lb ft3 A W S V [ft3] W [lb] A W S V [ft3] W [lb] VA,F WA,F VA,B WA,B in the borrow block diagram, and from there we can solve for the total volume of --- VW,F WW,F VW,B WW,B VS WS VS WS WT,F VT,B WT,B 10,000

wc=10% Borrow wc=10% S=50% SG=2.67 γdry=110 lb ft3 A W S V [ft3] W [lb] A W S V [ft3] W [lb] VA,F WA,F VA,B WA,B borrowed material. [pause] We’ll also assume the weight of air, W A, --- VW,F WW,F VW,B WW,B VS WS VS WS WT,F VT,B WT,B 10,000

wc=10% Borrow wc=10% S=50% SG=2.67 γdry=110 lb ft3 A W S V [ft3] W [lb] A W S V [ft3] W [lb] VA,F VA,B equals 0 pounds. [pause] The total weight of the fill material equals, --- VW,F WW,F VW,B WW,B VS WS VS WS WT,F VT,B WT,B 10,000

γdry=110 Find: VTotal[ft3] of borrowed soil A W S Fill VT=10,000 [ft3]
wc=10% WT,F = WW,F + WS γdry=110 lb ft3 V [ft3] W [lb] VA,F A the weight of the water, plus the weight solid material. Since the water content is defined as --- VW,F WW,F W S VS WS WT,F 10,000

wc=10% WT,F = WW,F + WS WW,F wc= γdry=110 WS lb ft3 V [ft3] W [lb] VA,F A the weight of the water divided by the weight of the solid material, we know the weight of the water equals, --- VW,F WW,F W S VS WS WT,F 10,000

wc=10% WT,F = WW,F + WS WW,F wc= γdry=110 WS lb ft3 WW,F = wc * WS V [ft3] W [lb] VA,F A the weight of the solid material times the water content. [pause] Making this substitution for the weight of the water ---- VW,F WW,F W S VS WS WT,F 10,000

wc=10% WT,F = WW,F + WS WW,F wc= γdry=110 WS lb ft3 WW,F = wc * WS V [ft3] W [lb] VA,F A into our previous equation, we find the weight of the water equals, --- VW,F WW,F W S VS WS WT,F 10,000

wc=10% WT,F = WW,F + WS WW,F wc= γdry=110 WS lb ft3 WW,F = wc * WS V [ft3] W [lb] WT,F = wc * WS +WS VA,F A the weight of solid material times the quantity, --- VW,F WW,F W S VS WS WT,F 10,000

wc=10% WT,F = WW,F + WS WW,F wc= γdry=110 WS lb ft3 WW,F = wc * WS V [ft3] W [lb] WT,F = wc * WS +WS VA,F A 1 plus the water content. Solving for the weight of the solid material, equals, --- VW,F WW,F WT,F = WS (1+wc) W S VS WS WT,F 10,000

wc=10% WT,F = WW,F + WS WW,F wc= γdry=110 WS lb ft3 WW,F = wc * WS V [ft3] W [lb] WT,F = wc * WS +WS VA,F A the total weight of the fill material, divided by, 1 plus the water content. The dry unit weight of a soil equals, --- VW,F WW,F WT,F = WS (1+wc) W S WT,F VS WS WS = (1+wc) WT,F WT,F 10,000

γdry= γdry=110 Find: VTotal[ft3] of borrowed soil A W S WT,F Fill
VT=10,000 [ft3] wc=10% WS = (1+wc) WS γdry= γdry=110 lb VT,F ft3 V [ft3] W [lb] VA,F A the weight of the solid material, divided by the total volume. Substituting in our value for the --- VW,F WW,F W S VS WS WT,F WT,F 10,000

γdry= γdry=110 Find: VTotal[ft3] of borrowed soil A W S WT,F Fill
VT=10,000 [ft3] wc=10% WS = (1+wc) WS γdry= γdry=110 lb VT,F ft3 V [ft3] W [lb] VA,F A weight of the solid material, ---- VW,F WW,F W S VS WS WT,F WT,F 10,000

γdry= γdry=110 γdry*VT,F= Find: VTotal[ft3] of borrowed soil A W S
WT,F Fill VT=10,000 [ft3] wc=10% WS = (1+wc) WS γdry= γdry=110 lb VT,F ft3 V [ft3] W [lb] WT,F γdry*VT,F= VA,F A (1+wc) solving for the total weight of the fill material, --- VW,F WW,F W S VS WS WT,F WT,F 10,000

WT,F Fill VT=10,000 [ft3] wc=10% WS = (1+wc) WS γdry= γdry=110 lb VT,F ft3 V [ft3] W [lb] WT,F γdry*VT,F= VA,F A (1+wc) and plugging in the known quantities, --- WT,F=γdry*VT,F*(1+wc) VW,F WW,F W S VS WS WT,F WT,F 10,000

WT,F Fill VT=10,000 [ft3] wc=10% WS = (1+wc) WS γdry= γdry=110 lb VT,F ft3 V [ft3] W [lb] WT,F γdry*VT,F= VA,F A (1+wc) the total weight of the fill material equals, --- WT,F=γdry*VT,F*(1+wc) VW,F WW,F W S VS WS WT,F WT,F 10,000

WT,F Fill VT=10,000 [ft3] wc=10% WS = (1+wc) WS γdry= γdry=110 lb VT,F ft3 V [ft3] W [lb] WT,F γdry*VT,F= VA,F A (1+wc) 1.21*106, pounds. Now we can solve for the weight of the solid material, by plugging in the --- WT,F=γdry*VT,F*(1+wc) VW,F WW,F W S VS WS WT,F = 1.21 * 106 [lb] WT,F 10,000

γdry=110 γdry*VT,F= Find: VTotal[ft3] of borrowed soil A W S WT,F Fill
VT=10,000 [ft3] wc=10% WS = (1+wc) γdry=110 lb ft3 V [ft3] W [lb] WT,F γdry*VT,F= VA,F A (1+wc) total weight and water content, the weight of the solid material equals, --- WT,F=γdry*VT,F*(1+wc) VW,F WW,F W S VS WS WT,F = 1.21 * 106 [lb] WT,F 10,000

γdry=110 γdry*VT,F= Find: VTotal[ft3] of borrowed soil A W S WT,F Fill
VT=10,000 [ft3] wc=10% WS = (1+wc) γdry=110 lb WS = 1.1*106 [lb] ft3 V [ft3] W [lb] WT,F γdry*VT,F= VA,F A (1+wc) 1.1*106 pounds. [pause] This value is added to the block diagram --- WT,F=γdry*VT,F*(1+wc) VW,F WW,F W S VS WS WT,F = 1.21 * 106 [lb] WT,F 10,000

γdry=110 Find: VTotal[ft3] of borrowed soil A W S WT,F Fill
VT=10,000 [ft3] wc=10% WS = (1+wc) γdry=110 lb WS = 1.1*106 [lb] ft3 V [ft3] W [lb] VA,F A and we know it is the same weight of solid material as the weight of solid material in the --- VW,F WW,F W S VS 1.1*106 WT,F 10,000

γdry=110 Find: VTotal[ft3] of borrowed soil A A W W S S WT,F Fill
VT=10,000 [ft3] wc=10% WS = (1+wc) γdry=110 lb WS = 1.1*106 [lb] ft3 V [ft3] W [lb] V [ft3] W [lb] VA,F VA,B A A block diagram for the borrowed material. [pause] To find the total volume of the borrowed material, --- VW,F WW,F VW,B WW,B W W S S VS 1.1*106 VS 1.1*106 WT,F VT,B WT,B 10,000

Find: VTotal[ft3] of borrowed soil
V [ft3] W [lb] VA,B wc=10% A S=50% VW,B WW,B W SG=2.67 S VS 1.1*106 VT,B=VA,B+VW,B+VS VT,B WT,B we’ll sum the volumes of air, water and solids. In this equation, the volumes of water and solid material equal, ---

γW γW*SG Find: VTotal[ft3] of borrowed soil A W S Borrow VA,B wc=10%
V [ft3] W [lb] VA,B wc=10% A S=50% VW,B WW,B W SG=2.67 S VS 1.1*106 VT,B=VA,B+VW,B+VS VT,B WT,B the weight divided by the unit weight. And the volume of air is a function of the volume of water and --- WW,B WS VW,B= VS= γW γW*SG

V [ft3] W [lb] VA,B wc=10% A S=50% VW,B WW,B W SG=2.67 S VS 1.1*106 VT,B=VA,B+VW,B+VS VT,B WT,B the saturation of the soil, S. [pause] From earlier, we determined the weight of water equals, --- WW,B WS VW,B= VS= γW γW*SG VW,B * (1-S) VA,B= S

V [ft3] W [lb] VA,B wc=10% A S=50% VW,B WW,B W SG=2.67 S VS 1.1*106 VT,B=VA,B+VW,B+VS VT,B WT,B the water content, times, weight of the solid material. After plugging in the known variables, --- WW,B WS VW,B= VS= γW γW*SG VW,B * (1-S) VA,B= WW,B = wc * WS S

V [ft3] W [lb] VA,B wc=10% A S=50% VW,B WW,B W SG=2.67 S VS 1.1*106 VT,B=VA,B+VW,B+VS VT,B WT,B the weight of water in the borrowed soil equals, --- WW,B WS VW,B= VS= γW γW*SG VW,B * (1-S) VA,B= WW,B = wc * WS S

V [ft3] W [lb] VA,B wc=10% A S=50% VW,B WW,B W SG=2.67 S VS 1.1*106 VT,B=VA,B+VW,B+VS VT,B WT,B 1.1*105 pounds. Next we’ll substitute in the weight of water and solid material, --- WW,B WS VW,B= VS= γW γW*SG VW,B * (1-S) VA,B= WW,B = wc * WS S WW,B = 1.1*105 [lb]

V [ft3] W [lb] VA,B wc=10% A S=50% VW,B W 1.1*105 SG=2.67 S VS 1.1*106 VT,B=VA,B+VW,B+VS VT,B WT,B as well as, the unit weight of water and ---- WW,B WS VW,B= VS= γW γW*SG VW,B * (1-S) VA,B= WW,B = wc * WS S WW,B = 1.1*105 [lb]

γW = 62.4 γW γW*SG Find: VTotal[ft3] of borrowed soil A W S Borrow
V [ft3] W [lb] VA,B wc=10% A γW = 62.4 lb S=50% VW,B W 1.1*105 ft3 SG=2.67 S VS 1.1*106 VT,B=VA,B+VW,B+VS VT,B WT,B specific gravity, and the volumes of water and solid material equal, --- WW,B WS VW,B= VS= γW γW*SG VW,B * (1-S) VA,B= S

γW = 62.4 γW γW*SG Find: VTotal[ft3] of borrowed soil A W S Borrow
V [ft3] W [lb] VA,B wc=10% A γW = 62.4 lb S=50% VW,B W 1.1*105 ft3 SG=2.67 S VS 1.1*106 VT,B=VA,B+VW,B+VS VT,B WT,B 1,763 cubic feet and 6,602 cubic feet, respectively. [pause] The volume of air is equal to --- WW,B WS VW,B= VS= γW γW*SG VW,B * (1-S) VW,B = 1,763 [ft3] VA,B= S VS = 6,602 [ft3]

V [ft3] W [lb] wc=10% A VA,B S=50% W 1,763 1.1*105 SG=2.67 S 6,602 1.1*106 VT,B=VA,B+VW,B+VS VT,B WT,B the volume of water, times, 1 minus the saturation as a decimal, divided by the saturation, as a decimal. VW,B * (1-S) VA,B= S

V [ft3] W [lb] wc=10% A VA,B S=50% 0.50 W 1,763 1.1*105 SG=2.67 S 6,602 1.1*106 VT,B=VA,B+VW,B+VS VT,B WT,B after plugging in the appropriate values, the volume of air equals, --- VW,B * (1-S) VA,B= S

V [ft3] W [lb] wc=10% A 1,763 S=50% 0.50 W 1,763 1.1*105 SG=2.67 S 6,602 1.1*106 VT,B=VA,B+VW,B+VS VT,B WT,B 1,763 cubic feet. [pause] This makes the total volume --- 1,763 [ft3] VW,B * (1-S) VA,B= S

V [ft3] W [lb] wc=10% A 1,763 S=50% 0.50 W 1,763 1.1*105 SG=2.67 S 6,602 1.1*106 VT,B=VA,B+VW,B+VS VT,B WT,B in the borrowed sample of soil equal to, --- 1,763 [ft3] VW,B * (1-S) VA,B= S

V [ft3] W [lb] wc=10% A 1,763 S=50% 0.50 W 1,763 1.1*105 SG=2.67 S 6,602 1.1*106 VT,B=VA,B+VW,B+VS VT,B WT,B VT,B=10,128 [ft3] 10,128 cubic feet. [pause] 1,763 [ft3] VW,B * (1-S) VA,B= S

V [ft3] W [lb] wc=10% A 1,763 S=50% 0.50 W 1,763 1.1*105 SG=2.67 S 6,602 1.1*106 VT,B=VA,B+VW,B+VS VT,B WT,B VT,B=10,128 [ft3] 371 9,872 10,000 10,128 When reviewing the possible solutions, --- 1,763 [ft3] VW,B * (1-S) VA,B= S

V [ft3] W [lb] wc=10% A 1,763 S=50% 0.50 W 1,763 1.1*105 SG=2.67 S 6,602 1.1*106 VT,B=VA,B+VW,B+VS VT,B WT,B VT,B=10,128 [ft3] 371 9,872 10,000 10,128 the answer is D. AnswerD VW,B * (1-S) VA,B= S

371 9,872 10,000 10,128 Some of the soil data is provided, for each site, as well as the specific gravity.

Some of the soil data is provided, for each site, as well as the specific gravity.

( ) ? 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

γdry=110 Find: VTotal[ft3] of borrowed soil Fill VT=10,000 [ft3]

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γdry=110 Find: VTotal[ft3] of borrowed soil Fill VT=10,000 [ft3]

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Presentation on theme: "γdry=110 Find: VTotal[ft3] of borrowed soil Fill VT=10,000 [ft3]"— Presentation transcript:

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