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Catégorie :Category: nCreator TI-Nspire
Auteur Author: brandonmervin
Type : Classeur 3.0.1
Page(s) : 1
Taille Size: 3.82 Ko KB
Mis en ligne Uploaded: 26/04/2025 - 08:39:31
Mis à jour Updated: 26/04/2025 - 08:49:25
Uploadeur Uploader: brandonmervin (Profil)
Téléchargements Downloads: 2
Visibilité Visibility: Archive publique
Shortlink : http://ti-pla.net/a4603327
Type : Classeur 3.0.1
Page(s) : 1
Taille Size: 3.82 Ko KB
Mis en ligne Uploaded: 26/04/2025 - 08:39:31
Mis à jour Updated: 26/04/2025 - 08:49:25
Uploadeur Uploader: brandonmervin (Profil)
Téléchargements Downloads: 2
Visibilité Visibility: Archive publique
Shortlink : http://ti-pla.net/a4603327
Description
Fichier Nspire généré sur TI-Planet.org.
Compatible OS 3.0 et ultérieurs.
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¡Claro! Vamos a resolver ambos ejercicios paso por paso. son el 1 y el 2 Primero te explico rápidamente las fórmulas que usaremos: Relación de vacíos e = V v V s e = frac{V_v}{V_s} e = V s V v Porosidad n = V v V t = e 1 + e n = frac{V_v}{V_t} = frac{e}{1+e} n = V t V v = 1 + e e Grado de saturación S r = V w V v = w G s e S_r = frac{V_w}{V_v} = frac{w G_s}{e} S r = V v V w = e w G s Densidad seca Á s = Á 1 + w rho_s = frac{rho}{1+w} Á s = 1 + w Á Peso unitario húmedo ³ = Á g gamma = rho g ³ = Á g Peso unitario seco ³ s = Á s g gamma_s = rho_s g ³ s = Á s g g H 9.81 m/s 2 g approx 9.81 , text{m/s}^2 g H 9.81 m/s 2 Primer ejercicio Datos: Muestra: Arcilla limosa gris M H = 126 kg M_H = 126 , text{kg} M H = 126 kg Á = 2.05 g/cm 3 = 2050 kg/m 3 rho = 2.05 , text{g/cm}^3 = 2050 , text{kg/m}^3 Á = 2.05 g/cm 3 = 2050 kg/m 3 G s = 2.71 G_s = 2.71 G s = 2.71 w = 15.7 % = 0.157 w = 15.7% = 0.157 w = 15.7% = 0.157 Calculamos: (a) Todos los valores del diagrama de fase Primero, densidad seca: Á s = Á 1 + w = 2050 1 + 0.157 H 1772 kg/m 3 rho_s = frac{rho}{1+w} = frac{2050}{1+0.157} approx 1772 , text{kg/m}^3 Á s = 1 + w Á = 1 + 0.157 2050 H 1772 kg/m 3 Relación de vacíos: e = G s Á w Á s 1 e = frac{G_s rho_w}{rho_s} - 1 e = Á s G s Á w 1 donde Á w = 1000 kg/m 3 rho_w = 1000 , text{kg/m}^3 Á w = 1000 kg/m 3 (densidad del agua). e = 2.71 × 1000 1772 1 H 0.529 e = frac{2.71 times 1000}{1772} - 1 approx 0.529 e = 1772 2.71 × 1000 1 H 0.529 Porosidad: n = e 1 + e = 0.529 1 + 0.529 H 0.346 ( 34.6 % ) n = frac{e}{1+e} = frac{0.529}{1+0.529} approx 0.346 , (34.6%) n = 1 + e e = 1 + 0.529 0.529 H 0.346 ( 34.6% ) Grado de saturación: S r = w G s e = 0.157 × 2.71 0.529 H 0.804 ( 80.4 % ) S_r = frac{w G_s}{e} = frac{0.157 times 2.71}{0.529} approx 0.804 , (80.4%) S r = e w G s = 0.529 0.157 × 2.71 H 0.804 ( 80.4% ) Peso unitario húmedo: ³ = Á g = 2050 × 9.81 H 20110.5 N/m 3 gamma = rho g = 2050 times 9.81 approx 20110.5 , text{N/m}^3 ³ = Á g = 2050 × 9.81 H 20110.5 N/m 3 Peso unitario seco: ³ s = Á s g = 1772 × 9.81 H 17378.5 N/m 3 gamma_s = rho_s g = 1772 times 9.81 approx 17378.5 , text{N/m}^3 ³ s = Á s g = 1772 × 9.81 H 17378.5 N/m 3 Segundo ejercicio Datos: Muestra: Cubo de 76 mm de lado Á = 1.44 g/cm 3 = 1440 kg/m 3 rho = 1.44 , text{g/cm}^3 = 1440 , text{kg/m}^3 Á = 1.44 g/cm 3 = 1440 kg/m 3 G s = 2.55 G_s = 2.55 G s = 2.55 w = 68.5 % = 0.685 w = 68.5% = 0.685 w = 68.5% = 0.685 Volumen del cubo: V = ( 0.076 m ) 3 = 0.000438976 m 3 V = (0.076 , text{m})^3 = 0.000438976 , text{m}^3 V = ( 0.076 m ) 3 = 0.000438976 m 3 (a) Relación de vacíos Primero, densidad seca: Á s = Á 1 + w = 1440 1 + 0.685 H 854.9 kg/m 3 rho_s = frac{rho}{1+w} = frac{1440}{1+0.685} approx 854.9 , text{kg/m}^3 Á s = 1 + w Á = 1 + 0.685 1440 H 854.9 kg/m 3 Relación de vacíos: e = G s Á w Á s 1 = 2.55 × 1000 854.9 1 H 1.984 e = frac{G_s rho_w}{rho_s} - 1 = frac{2.55 times 1000}{854.9} - 1 approx 1.984 e = Á s G s Á w 1 = 854.9 2.55 × 1000 1 H 1.984 (b) Grado de saturación S r = w G s e = 0.685 × 2.55 1.984 H 0.880 ( 88.0 % ) S_r = frac{w G_s}{e} = frac{0.685 times 2.55}{1.984} approx 0.880 , (88.0%) S r = e w G s = 1.984 0.685 × 2.55 H 0.880 ( 88.0% ) (c) Peso total húmedo M H = Á V = 1440 × 0.000438976 H 0.6321 kg M_H = rho V = 1440 times 0.000438976 approx 0.6321 , text{kg} M H = Á V = 1440 × 0.000438976 H 0.6321 kg (d) Contenido de humedad si el suelo se satura ( S r = 1 S_r = 1 S r = 1 ) Sabemos que: w s a t = e G s w_{sat} = frac{e}{G_s} w s a t = G s e w s a t = 1.984 2.55 H 0.778 ( 77.8 % ) w_{sat} = frac{1.984}{2.55} approx 0.778 , (77.8%) w s a t = 2.55 1.984 H 0.778 ( 77.8% ) (e) Todos los valores del diagrama de fase Ya calculamos: e = 1.984 e = 1.984 e = 1.984 S r = 88.0 % S_r = 88.0% S r = 88.0% w = 68.5 % w = 68.5% w = 68.5% w s a t = 77.8 % w_{sat} = 77.8% w s a t = 77.8% Á s = 854.9 kg/m 3 rho_s = 854.9 , text{kg/m}^3 Á s = 854.9 kg/m 3 (f) Densidad seca sin cambio en volumen total Ya la tenemos: Á s H 854.9 kg/m 3 rho_s approx 854.9 , text{kg/m}^3 Á s H 854.9 kg/m 3 Made with nCreator - tiplanet.org
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Compatible OS 3.0 et ultérieurs.
<<
¡Claro! Vamos a resolver ambos ejercicios paso por paso. son el 1 y el 2 Primero te explico rápidamente las fórmulas que usaremos: Relación de vacíos e = V v V s e = frac{V_v}{V_s} e = V s V v Porosidad n = V v V t = e 1 + e n = frac{V_v}{V_t} = frac{e}{1+e} n = V t V v = 1 + e e Grado de saturación S r = V w V v = w G s e S_r = frac{V_w}{V_v} = frac{w G_s}{e} S r = V v V w = e w G s Densidad seca Á s = Á 1 + w rho_s = frac{rho}{1+w} Á s = 1 + w Á Peso unitario húmedo ³ = Á g gamma = rho g ³ = Á g Peso unitario seco ³ s = Á s g gamma_s = rho_s g ³ s = Á s g g H 9.81 m/s 2 g approx 9.81 , text{m/s}^2 g H 9.81 m/s 2 Primer ejercicio Datos: Muestra: Arcilla limosa gris M H = 126 kg M_H = 126 , text{kg} M H = 126 kg Á = 2.05 g/cm 3 = 2050 kg/m 3 rho = 2.05 , text{g/cm}^3 = 2050 , text{kg/m}^3 Á = 2.05 g/cm 3 = 2050 kg/m 3 G s = 2.71 G_s = 2.71 G s = 2.71 w = 15.7 % = 0.157 w = 15.7% = 0.157 w = 15.7% = 0.157 Calculamos: (a) Todos los valores del diagrama de fase Primero, densidad seca: Á s = Á 1 + w = 2050 1 + 0.157 H 1772 kg/m 3 rho_s = frac{rho}{1+w} = frac{2050}{1+0.157} approx 1772 , text{kg/m}^3 Á s = 1 + w Á = 1 + 0.157 2050 H 1772 kg/m 3 Relación de vacíos: e = G s Á w Á s 1 e = frac{G_s rho_w}{rho_s} - 1 e = Á s G s Á w 1 donde Á w = 1000 kg/m 3 rho_w = 1000 , text{kg/m}^3 Á w = 1000 kg/m 3 (densidad del agua). e = 2.71 × 1000 1772 1 H 0.529 e = frac{2.71 times 1000}{1772} - 1 approx 0.529 e = 1772 2.71 × 1000 1 H 0.529 Porosidad: n = e 1 + e = 0.529 1 + 0.529 H 0.346 ( 34.6 % ) n = frac{e}{1+e} = frac{0.529}{1+0.529} approx 0.346 , (34.6%) n = 1 + e e = 1 + 0.529 0.529 H 0.346 ( 34.6% ) Grado de saturación: S r = w G s e = 0.157 × 2.71 0.529 H 0.804 ( 80.4 % ) S_r = frac{w G_s}{e} = frac{0.157 times 2.71}{0.529} approx 0.804 , (80.4%) S r = e w G s = 0.529 0.157 × 2.71 H 0.804 ( 80.4% ) Peso unitario húmedo: ³ = Á g = 2050 × 9.81 H 20110.5 N/m 3 gamma = rho g = 2050 times 9.81 approx 20110.5 , text{N/m}^3 ³ = Á g = 2050 × 9.81 H 20110.5 N/m 3 Peso unitario seco: ³ s = Á s g = 1772 × 9.81 H 17378.5 N/m 3 gamma_s = rho_s g = 1772 times 9.81 approx 17378.5 , text{N/m}^3 ³ s = Á s g = 1772 × 9.81 H 17378.5 N/m 3 Segundo ejercicio Datos: Muestra: Cubo de 76 mm de lado Á = 1.44 g/cm 3 = 1440 kg/m 3 rho = 1.44 , text{g/cm}^3 = 1440 , text{kg/m}^3 Á = 1.44 g/cm 3 = 1440 kg/m 3 G s = 2.55 G_s = 2.55 G s = 2.55 w = 68.5 % = 0.685 w = 68.5% = 0.685 w = 68.5% = 0.685 Volumen del cubo: V = ( 0.076 m ) 3 = 0.000438976 m 3 V = (0.076 , text{m})^3 = 0.000438976 , text{m}^3 V = ( 0.076 m ) 3 = 0.000438976 m 3 (a) Relación de vacíos Primero, densidad seca: Á s = Á 1 + w = 1440 1 + 0.685 H 854.9 kg/m 3 rho_s = frac{rho}{1+w} = frac{1440}{1+0.685} approx 854.9 , text{kg/m}^3 Á s = 1 + w Á = 1 + 0.685 1440 H 854.9 kg/m 3 Relación de vacíos: e = G s Á w Á s 1 = 2.55 × 1000 854.9 1 H 1.984 e = frac{G_s rho_w}{rho_s} - 1 = frac{2.55 times 1000}{854.9} - 1 approx 1.984 e = Á s G s Á w 1 = 854.9 2.55 × 1000 1 H 1.984 (b) Grado de saturación S r = w G s e = 0.685 × 2.55 1.984 H 0.880 ( 88.0 % ) S_r = frac{w G_s}{e} = frac{0.685 times 2.55}{1.984} approx 0.880 , (88.0%) S r = e w G s = 1.984 0.685 × 2.55 H 0.880 ( 88.0% ) (c) Peso total húmedo M H = Á V = 1440 × 0.000438976 H 0.6321 kg M_H = rho V = 1440 times 0.000438976 approx 0.6321 , text{kg} M H = Á V = 1440 × 0.000438976 H 0.6321 kg (d) Contenido de humedad si el suelo se satura ( S r = 1 S_r = 1 S r = 1 ) Sabemos que: w s a t = e G s w_{sat} = frac{e}{G_s} w s a t = G s e w s a t = 1.984 2.55 H 0.778 ( 77.8 % ) w_{sat} = frac{1.984}{2.55} approx 0.778 , (77.8%) w s a t = 2.55 1.984 H 0.778 ( 77.8% ) (e) Todos los valores del diagrama de fase Ya calculamos: e = 1.984 e = 1.984 e = 1.984 S r = 88.0 % S_r = 88.0% S r = 88.0% w = 68.5 % w = 68.5% w = 68.5% w s a t = 77.8 % w_{sat} = 77.8% w s a t = 77.8% Á s = 854.9 kg/m 3 rho_s = 854.9 , text{kg/m}^3 Á s = 854.9 kg/m 3 (f) Densidad seca sin cambio en volumen total Ya la tenemos: Á s H 854.9 kg/m 3 rho_s approx 854.9 , text{kg/m}^3 Á s H 854.9 kg/m 3 Made with nCreator - tiplanet.org
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