Indices Fisicos - Jaque e Sarah

PUC Minas 21/09 Lista de Exercício - Indice físicos RA: 667860 ; 637009 \rightarrow N=5 D) a) V = 5,98 \times 10^{-4} \ m^3 \ m_s: 0,918_g \ g=10 \ m/s^2 m=1,010_{kg} \gamma=26,5 \ kN/m^3 e=? \ \quad \ e = \frac{V_v}{V_s} \varepsilon=26,5 \times 10^3: 0,918 \times 10 \V_s 26,500 \ V_s:9,18 \quad \ V_s = 3,46 \times 10^{-4} V_v: V-Vs \quad V_v:2,52 \times 10^{-4} \frac{e = V_v} \quad \varepsilon \rightarrow \frac{2,52 \times 10^{-4}}{3,46 \times 10^{-4}} \rightarrow 0,72\eta e) \eta = \frac{V_v}{V} \rightarrow \frac{2,52 \times 10^{-4} \times 100}{5,98 \times 10^{-4}} \quad \langle \eta \quad = 42,19% \rangle c) \varphi = \frac{M-M_s}{M_s}, M_{u} = 97_g W_u: \frac{97 \times 100}{918} \rightarrow \epsilon \to \ L = 10,02 % W_u \to \ 10,07: \ int \ W_u = 10,02 % d) S_f = \frac{V_{uw}}{V_v} \ V_{uw} = \mu \frac{9,2 \times 10^{-5}}{3,6} \ cm^{3} de \ S_v = 9,2 \ times 10^{-5}\rightarrow 57 \ S_f = 36,51% e) \gamma = \frac{M}{V} \rightarrow \frac{1010_{kg} \times 10}{5,98 \times 10^{-4} \ m^3} \rightarrow 16,89_{kN}/m^3 g=10 \ m/s^2 2 a) \gamma_s=19,5_{kN}/m^3 \gamma_s=26,9 \ kN/m^3 w = 29 % \frac{\chi_d}{\chi_d} = \frac{X}{1+w} \rightarrow \frac{19,5}{1+0,29} = 15,12 \ kN/m^3 b) e = \frac{\chi_s (1+w) -1}{\chi} \rightarrow \frac{26,9(1+0,29)-1}{19,5} \left[\frac{26,9+7,801)-1}{19,5} \rightarrow 0,779 \right] c) \eta = \frac{e}{1+e} \rightarrow 0,779 \times 100 = \langle 43,79 % \rangle d) S_r = \frac{w \times \chi_s}{e \times \chi_w} \rightarrow \frac{0,29 \times 26,9}{0,779 \times (1 \times 0)} \times 100 \rightarrow S_r = 100,14 % 03. a) peso especifico natural x: M = 96 . 1,20 g/cm3 = 12,9 kN/m3 V 19 b) peso especifico aparente seco: xd = MS . 31 = 1,63g/cm3 = 16,3 kN/m3 V 19 c) peso especifico saturado: Mw = M-MS = 96-31 = 5g xw= Mw 1 = 5 = Vu: 6 cm3 Vw Sr: Nw = o 9,71o= o = Vu:6,71 cm3 Vw Vw xsat: Mg = (Vw.Xw) o o+ (6,71.1) = 1,98 g/cm3 V 19 d) indice de vazios: vo : V-Vw +Vo:19-6,71 =12,29 cm3 e: = Vv + e = 6,71 = o,546 V S 12,29 e) peso especifico dos solidos: xd: MS + 31 = 2,52 g/cm3 Vs 12,29 04. a) teor de umidade: xd . x -> 18,0 = 0,1818 -> w 1 + e = 21,9 -> w 1 + e = 1,4823 -> 1+w 1-w 18,5 w = 0,1783 = 17,38 % b) grau de saturacao: e: x8 = 1 -> 0,48+1 = x8 -> 1/42 18,5 = x8 -> x8 = 21,38kN/m3 xd Sr = w . xd = Sr = 0,4138 . 27,38 = 0,8013 = 89,131% Xw . e 10.048 c) peso especifico saturado: ysat: xs+e.xw= 23,38 . (0,48.10) = 21,74 kN/m3 1+e 1.048 d) peso especifico dos solidos: xs = 23,38 kN/m3 5) y: 20,5xN w=23% a) xd: x -> 20,5 -> yd = 16,67xN/m3 1+um 140,23 b) e: ? e: ydxum sr=100% (Sr.xwl)-(y.d.wl) Xwu=10 e: 16,67x0,23 -> 3,8341 e=0,622 (1y.10)-(16,67x0,23) 6,1659 c) ys=? ys: yd(1+e) ys=16,67(1+0,622) ys:16,67x1,622 ys=27,04 N/m3 6) M=95g V=50m3 MS=7,5g ys=26,5xN/m3 a) w=? Mw=M-Ms Mw=95-75 Mw=20 w=Mu x 100 w=20 x 100 w=26,67% MS 75 b) c=? e=ys-1 yd=y yd 1+w y=95x10 y=19kN yd=19 50 1+0,26 c=26,5-1 c=0,767 15 c) η:? η = e 1+e η = 0,767 x 100 = 43,41% 1+0,767 d) Se:?, Vuu = 100 Vv = mu -> mu = 20 g Vuu = 20 cm³ Vs= V x V 100 Vv = 21,71 se: = 20 x 100 21,71 Se: 92,12% e) γ:? γ = m x g v γ = 95 x 10 50 γ = 19 kN/m³ f) γd: ? 1+vu γd = 19 1+0,267 γd = 15 kN/m³ a) Indice de vazios: • mu = m total - m conjunto seco = 69,920 - 62,011 = 6,918 g • ms = m conjunto seco - recipiente = 62,011 - 35,046 = 26,965 g • Vs = ms = (26,965 / 2,85) x 8 = 9,46 cm³ ps (x/cm³) • M = ms + mu = 26,965 + 6,918 = 33,913 g • ρ = M = 2,85 = 33,913 -> V = 33,913 = 11,89 cm³ V x x 2,85 • Vv = V - Vs = 11,89 - 9,46 = 2,43 cm³ e = 2,43 = 0,25 9,46 (e) b) Teor de umidade: w = mu = 6,918 = 0,0553 = b = 5,53 % ms 26,965 c) Porosidade: n = Vv = 2,43 = 0,2044 = b = 20,44% V 11,89 8. Vz = 1000 m³ e = 1,20 • e' = Vv = 1,25 x 1000 = 1,9 = Ve = 785,56 m³ Vo Vse • e'' = Vv = 0,35 = Vv = 1 = 13,9372 (?) m³ de areia Vo 785,56 e:? 9) η:? V- D. Vs2 = π x D. Vs= 0,524D. 6 Vt = V - Vs e = Vv e = 0,476D. c = 0,91 Vs 0,524D. η = e 1+e η: 0,91 x 100 1+0,91 η = 47,64%! 10. r = 5 cm e = 1,0 e': 0,6 V = 4πr. V = 4π (10). = 523,60 cm. 3 3 c = Vv 1 = eVv + e = 523,60 -> Vø = 261,8 cm³ Vs 523,60 Vs e' = Vs -> 0,6 = Vs -> 1 = 0,6 x 261,8 = Vv = 418,88 cm³ Vo Vø e:? O novo volume da esfera é 418,88 cm³ 11) γs= 20kN/m^3 n=2,2 cm γS= 27,5kN/m^3 ν=4π x 3/3 V=4πx(0,022)^3/3 V=4,1602x10^-5m^3 a) w=7 w=γS-γ/γS w=27,5-20/27,5 x 100 w=27,27% b) e= χS (1+w)-1 e=27,5 (1+0,272)-1 χ χ e=0,749 c) η= e/1+e η=0,749 x100 η=47,82 1+0,749 d) γd: 1 = γd: 20 γd: 15,7kN/m^3 1+w 1+0,272