Obras de terra- barragens-estabilidade de taludes

terra homogênea com filtro horizontal, trincheira e reservatório cheio. 3- Foram utilizados como dados de entrada ao programa Slide, os seguintes parâmetros oriundos da amostra de solo ensaiada pelo CIENTEC, inicialmente para a determinação da linha freática, e após, para o cálculo de estabilidade dos taludes da seção modelada, em diferentes situações: 3.1 Peso específico saturado: γsat = γágua x n + γd Massa específica saturada = 1000 x 0.429 + 1539* = 1968 kg/m³ Peso específico saturado = 1968 x 9,81 = 19,306 kN/m³ ** * Massa específica seca com 95% de GC. ** Considerou-se como saturado todo o corpo do maciço para simular influência de chuva intensa ocorrendo simultaneamente na situação de reservatório cheio, sem se considerar, contudo, por esta circunstância, qualquer mudança ou alteração da linha freática originada pela percolação através do maciço da água do reservatório na sua cota máxima normal. 3.2 Solo da barragem: Coesão = 24, 0 KPa Ângulo de Atrito = 17.9° Permeabilidade (ky) = 8,55 x 10 -7 cm/s = 8,55 x 10 -9 m/s (unidade utilizada pelo programa Slide) Kx = 9Ky = 7,69 x 10 -8 m/s (o fator 9 foi adotado. Ver item 4). 3.3 Para o filtro de areia horizontal (Cruz, 1996): Coesão: zero Ângulo de atrito: 32° Permeabilidade: kx = ky = 3 x 10 -4 m/s 4- Linha Freática: Considerando-se o…

(b) The magnitude of the (only) force on particle 1 is q q q F = m1a1 = k 1 2 2 = 8.99 × 109 . r 0.0032 2 2 c h Inserting the values for m1 and a1 (see part (a)) we obtain |q| = 7.1 × 10–11 C. 2. The magnitude of the mutual force of attraction at r = 0.120 m is 3.00 × 10−6 150 × 10−6 . q1 q2 9 F=k = 8.99 × 10 = 2.81 N . 2 r 01202 . c hc hc h 3. Eq. 21-1 gives Coulomb’s Law, F = k k | q1 || q2 | r= = F q1 q2 r2 , which we solve for the distance:…

(b) The magnitude of the (only) force on particle 1 is q q q F = m1a1 = k 1 2 2 = 8.99 × 109 . r 0.0032 2 2 c h Inserting the values for m1 and a1 (see part (a)) we obtain |q| = 7.1 × 10–11 C. 2. The magnitude of the mutual force of attraction at r = 0.120 m is 3.00 × 10−6 150 × 10−6 . q1 q2 9 F=k = 8.99 × 10 = 2.81 N . 2 r 01202 . c hc hc h 3. Eq. 21-1 gives Coulomb’s Law, F = k k | q1 || q2 | r= = F q1 q2 r2 , which we solve for the distance:…

- TiplrChapter 4 Newton’s Laws Conceptual Problems 1 • While on a very smooth level transcontinental plane flight, your coffee cup sits motionless on your tray. Are there forces acting on the cup? If so, how do they differ from the forces that would be acting on the cup if it sat on your kitchen table at home? Determine the Concept Yes, there are forces acting on it. They are the normal force of the table and the gravitational pull of Earth (weight). Because the cup is not accelerating…

N 2. The magnitude of the mutual force of attraction at r = 0.120 m is F =k |q1 ||q2 | = 8.99 × 109 r2 3.00 × 10−6 1.50 × 10−6 = 2.81 N . 0.1202 3. (a) With a understood to mean the magnitude of acceleration, Newton’s second and third laws lead to m2 a2 = m1 a1 =⇒ m2 = 6.3 × 10−7 kg 7.0 m/s2 = 4.9 × 10−7 kg . 9.0 m/s 2 (b) The magnitude of the (only) force on particle 1 is F = m 1 a1 = k |q |2 |q1 ||q2 | = 8.99 × 109 . r2 0.00322 Inserting the…

have [pic] (d) Similarly, ycog = [0 + (2.00)(m)(7.80) + (4.00)(m)(7.60) + (4.00)(m)(7.40) + (2.00)(m)(7.60) + 0]/(8.00m + 7.80m + 7.60m + 7.40m + 7.60m + 7.80m) = 1.97 m. 2. The situation is somewhat similar to that depicted for problem 10 (see the figure that accompanies that problem). By analyzing the forces at the “kink” where [pic] is exerted, we find (since the acceleration is zero) 2T sin θ = F, where θ is the angle (taken positive) between each segment of the string and…

Machining Dynamics Tony L. Schmitz l Kevin S. Smith Machining Dynamics Frequency Response to Improved Productivity 13 Tony L. Schmitz University of Florida Mechanical and Aerospace Engineering Gainesville, FL 32611-6250 tschmitz@ufl.edu Kevin S. Smith University of North Carolina Mechanical Engineering and Engineering Science Dept. Charlotte, NC 28223-0001 kssmith@uncc.edu ISBN 978-0-387-09644-5 e-ISBN 978-0-387-09645-2 DOI 10.1007/978-0-387-09645-2 Library of…

have [pic] (d) Similarly, ycog = [0 + (2.00)(m)(7.80) + (4.00)(m)(7.60) + (4.00)(m)(7.40) + (2.00)(m)(7.60) + 0]/(8.00m + 7.80m + 7.60m + 7.40m + 7.60m + 7.80m) = 1.97 m. 2. The situation is somewhat similar to that depicted for problem 10 (see the figure that accompanies that problem). By analyzing the forces at the “kink” where [pic] is exerted, we find (since the acceleration is zero) 2T sin θ = F, where θ is the angle (taken positive) between each segment of the string and…

have [pic] (d) Similarly, ycog = [0 + (2.00)(m)(7.80) + (4.00)(m)(7.60) + (4.00)(m)(7.40) + (2.00)(m)(7.60) + 0]/(8.00m + 7.80m + 7.60m + 7.40m + 7.60m + 7.80m) = 1.97 m. 2. The situation is somewhat similar to that depicted for problem 10 (see the figure that accompanies that problem). By analyzing the forces at the “kink” where [pic] is exerted, we find (since the acceleration is zero) 2T sin θ = F, where θ is the angle (taken positive) between each segment of the string and…

have [pic] (d) Similarly, ycog = [0 + (2.00)(m)(7.80) + (4.00)(m)(7.60) + (4.00)(m)(7.40) + (2.00)(m)(7.60) + 0]/(8.00m + 7.80m + 7.60m + 7.40m + 7.60m + 7.80m) = 1.97 m. 2. The situation is somewhat similar to that depicted for problem 10 (see the figure that accompanies that problem). By analyzing the forces at the “kink” where [pic] is exerted, we find (since the acceleration is zero) 2T sin θ = F, where θ is the angle (taken positive) between each segment of the string and…