Indices Fisicos - Jaque e Sarah

PUC Minas 21/09 Lista de Exercicios - Indice fisico RA: 667860, 637009 -> N: 5 (a) V = 5,98 x 10**-4 m**3 m = 1,010 kg, ms = 0,918 kg, g = 10 m/s**2 Y = 26,5 kN/m**3 p = ? e = Vv/Vs 26,5 x 10**3 = 0,918 x 10 26.500 = Vs x 9,18 Vs = 3,46 x 10**-4 Vv = V - Vs Vv = 2,52 x 10**-4 e = 2,52x10**-4 3,46 x 10**-4 p = 0,728 (b) eta = Vv/V -> 2,52 x 10**-4 x 100/5,98 x 10**-4 eta = 42,14 % (c) W = Mu/Ms Mu_i = M - Ms Mu_i = 92 g W = 92 x 100/918 W = 10,02 % (d) Sr = Vuu/Vv Vuu = muu -> 92 cm**3 9,2 x 10**-5 m**3 Sr = 9,2 x 10**-5 x 100/2,52 x 10**-4 Sr = 36,51 % e) g = M/V g = 1010 xg x 10 5,98 x 10**-4 m**3 16,89 k.N/m**3 Y = 10 m/s**2 2) a) Y = 19,5 kN/m**3 g_s = 26,9 kN/m**3 W = 29 % X_d = X/i+u -> 19,5 = 1.5,12 kN/m**3 1+0,29 (b) e = Xs(1+u)-1 Y e 26,9(1+0,29)-1 19,5 26,9 x (7,801)-1 -> 0,779 19,5 (c) eta = e/1+e -> 0,779 x 100/1+0,779 eta = 43,39 % (d) Sr = u x g_s -> 0,29 x 26,9 x 100/ 0,779 x (1 x 10) e x W Sr = 100,14 % 03. a) peso especifico natural \[X = \frac{M}{V} = \frac{86}{19} = 1,29 \ g/cm^3 = 12,9 \ kN/m^3\] b) peso especifico aparente seco: \[Xd = \frac{Ms}{V} = \frac{31}{19} = 1,63 \ g/cm^3 = 16,3 \ kN/m^3\] c) peso especifico saturado: \[Mws = M-Ms = 86-31 = 5g \] \[Xw = \frac{Mws}{V} = \frac{5}{5} = \frac{Vws}{Vws} : Vws: 8cm^3\] \[Sr = \frac{Vw}{V} = \frac{0}{10+6} = 6 \] \[Vv = 6,71cm^3\] \[Xsat = \frac{Ms + (Vv . Xw)}{V} = \frac{81 + (6,71 . 1 )}{19} = 1,98 \ g/cm^3\] d) indice de vazios: \[Ve = Vv - Vv \rightarrow Ve: 19 - 6,71 = 12,29 \ cm^3\] \[e = \frac{Vv}{Vs} = \frac{6,71}{12,29} = 0,546 \] e) peso especifico dos sólidos: \[Xs = \frac{Ms}{Vs} = \frac{31}{12,29} = 2,52 \ g/cm^3\] 04. a) teor de umidade: \[Xd : X \rightarrow 18,0 = 0,1785 \] \[X = 0,1785\] \[W = 0,1785 \rightarrow 19,38 \]\] b) grau de saturação: \[e: Xs = 1 \rightarrow 0,418 + 1 = Xs\] \[\frac{1}{242} = 18,6 = \frac{Xs}{16,8}\] \[85r = \frac{w . Xs}{Xw.e} = \frac{0,418}{1} = 0,801428 . 20213\] c) peso especifico saturado: \[Xsat : Xs + e . Xw : 23,58 = (0,48 . 10) = 0.74 \frac{kN}{m^3}\] d) peso especifico dos sólidos: \[Xs = 22,92 . \frac{kN}{m^3}\] 5) Y: 20,5 x N\nu: 23% a) Xd : X \rightarrow 20,5 = \frac{{Yd}}{{1+\nu 0,23}} Yd = 16,67 x \frac{kN}{m^3} b) e: 7 \rightarrow e \frac{{Yd x \nu}}{{\frac{{Sr.xyw.(\frac{{yd.xuw}}{\lu)}}}{10}}} \[C = 16,67 x 0,23\] \rightarrow 3,8341 \rightarrow e: 0.622 Ys = 16.67x {1+ e)}\ Ys: 16,67x(1+0.622)\ Ys: 27,04\frac{kN}{m^3} 6) M: 95g \nu: 50m^3 Y = X/Y.V Mu: M - Ms. Ms: 75 g Ys = 26,5 . \frac{{kN}}{m^3} \[e: Ys - \frac{e}{1+nu}\] \[\ = \frac{(e.xy)}{10}^+ (90,23)\] 9) e: ? n2: ? V - D³ -> V52 = D / Vs = 0,524 D³ 6 x D³ V1 = V - Vs -> D³ - 0,524 D³ + Vv = 0,476D³ v = Vv e = 0,476 D³ e = 0,91 Vs = 0,524 D³ e = e n = 0,91 x 100 n = 47,64% 1 + c 1 + 0,91 10. r = 5cm e = 1,0 e' = 0,6 .v = 4 pi r³ ~ V = 4π(10)² = 523,60 cm² 3 3 .v' = Vt + Vv e = 1 = Vv + 2 = 523,60 -> Vb = 261,8 cm³ Vb Vb Vb 261,8 .e' = Vy -> 0,16 = Vy - 1 = 1,6 - 261.8 = Vv = 418,88 cm³ Vb 261,8 .N. O novo é volume da esfera é 418,88 cm³ c) n2: ? n = e n = 0,767 x 100 n = 43,41%. 1 + e 1 + 0,767 d) Se e Vv = 100 Vvu = mu = mu = 20g Vv Vvu = 20 cm³ Vt = V x V Vt = 21,71 Se = 20 x 100 Se = 32,12% Vt 21,71 e) g: ? g = m x g g = 35 x 10 g = 19 kN/m³ V 50 f) gd: X gd = 19 gd = 15 x N/m³ 1 + u 1 + 0,267 011 a) Indice de vazios: . Mw = Mtotal - Mconjuntoseco = 69,989 - 62,011 = 6,918g . Ma = Mconjuntoseco . MAciendente = 62,011 - 35,046 = 26,965g . Vs : Ms = 26,965 x 9,46 cm³ ps 2,85 cm³ .M = Ms + Mw = 26,965 + 6,918 = 33,913g .p = M = 2,85 = 33,913 -> V = 33,913 = 11,889 cm³ V 28,8 %V% = V - Vs = 11,89 - 9,46 = 2,43 cm³ e = 2,43 = 0,25 V 9,46 b) teor de umidade: .w = Mw = 6,918 = 0,25733 b 23,33% Mo 26,965 c). porosidade: .n = V = 2,43 = 0,2044 b 20,44 % V 11,89 08. Vz = 1000 m³ e = 1,20 .e' = Vy -> | 1,25 * 1000 - 1 | Vb = 7855,56 m³ Vo Vb .e'' = Vy | Vv -> 0,35 = Vv - 1 = 139,74,7 m³ de acima Vb Vb 955,55 11) γs= 20 kN/m³ n = 2,2 cm γsc= 27,5 kN/m³ γc= 27,5 kN/m³ Vs= 4π x (l/3)³ Vs= 4π x (0,022)³ /3 V= 4,160 x 10^-5 m³ a) w= ? w= γsc - γs γs w= 27,5 - 20 x100 27,5 [w = 27,3%] b) e= Xs (1+ w) -1 → e= 27,5 (1+ 0,272) -1 γ γs z0 [e = 0,749] c) η= e 1+e η= 0,749 x100 → η= 1+ 0,749 [y= 47,82] d) γd= γ 1+w γd= 20 1+w 1+ 0,272 1+w [γd= 15,7 kN/m³]