# Estatika

Solo disponible en BuenasTareas
• Páginas : 6 (1304 palabras )
• Descarga(s) : 0
• Publicado : 17 de junio de 2011

Vista previa del texto
CALCULOS Y RESULTADOS

TRABAJO DE GABINETE

1. Error de Cierre : Error de cierre = cota retorno – cota de ida
Error de cierre = ∑ vistas atrás - ∑ vistas adelante.

Error de cierre = cota retorno – cota de inicio
Ec = 50.303m – 50.00m
Ec = 0.303m

Error de cierre = ∑ vistas atrás - ∑ vistas adelante.
Ec =37.824-37.521
Ec = 0.303

2. Compensación delError de cierre

C= x dist. Parcial o acumulada
Dist. Total = 220m
C(B) = x20 = 0.0275m

C(C) = x40 = 0.0550m

C(D) = x60 = 0.0826m

C(E) = x80 = 0.1101m

C(F) = x100 = 0.1377m

C(G) = x120= 0.1652m

C(H) = x140 = 0.1928m

C(I) = x160 =0.2203m

C(J) = x180 = 0.2479m

C(K) = x200 = 0.2754m

C(L) = x220 = 0.303m

C(A) = x0 = 0m

3. Compensar cotas de ida yretorno

Cotas compensadas en Ida:

C(A) = 50.195 - 0 = 50.195m
C(B) = 50.476 - 0.0275 = 50.4485m
C(C) = 50.969 - 0.0550 = 50.914m
C(D) = 50.443 - 0.0826 = 50.3604m
C(E) = 50.219 - 0.1101 = 50.1089m
C(F) = 51.119 - 0.1377 = 50.9813m
C(G) = 51.115 - 0.1652 = 50.9498m
C(H) = 51.362 - 0.1928 = 51.1692m
C(I) = 49.893 - 0.2203 = 49.6727m
C(J) = 50.141- 0.2479 = 49.8931m
C(K) = 50.657 - 0.2754 = 50.3816m
C(L) = 51.334 - 0.303 = 51.031m

Cotas compensadas en Retorno:

C(L) = 51.334 - 0 = 51.334m
C(K) = 50.667 - 0.0275 = 50.6395m
C(J) = 50.189 - 0.0550 = 50.13m
C(I) = 49.956 - 0.0826 = 49.8734m
C(H) = 51.438- 0.1101 = 51.3279m
C(G) = 51.231 - 0.1377 = 51.0933m
C(F) = 51.207 - 0.1652 = 51.0418m
C(E) = 50.318 - 0.1928 = 50.1252m
C(D) = 50.532 - 0.2203 = 50.3117m
C(C) = 51.075 - 0.2479 = 50.8271m
C(B) = 50.572 - 0.2754 = 50.2966m
C(A) = 50.303 - 0.303 = 50m

4. Cotas promedio

(0)A = (50.195+50)/2 =50.0975
(20)B = (50.2966+50.4485)/2 =50.37255
(40)C =(50.8271+50.914)/2 =50.87055
(60)D = (50.3117+50.3604)/2 =50.33605
(80)E = (50.1252+50.1089)/2 =50.11705
(100)F = (51.0418+50.9813)/2 =51.01155
(120)G = (51.0933+50.9498)2 =51.02155
(140)H = (51.3279+51.1692)/2 =51.24855
(160)I = (49.8734+49.6727)/2 =49.77305
(180)J= (50.134+49.8931)/2 =50.01355
(200)K= (50.6395+50.3816)/2 =50.51055
(220)L= (51.334+51.031)/2=51.1825

5. Calculo de la Pendiente Natural del terreno

S = cota final - cota inicial

S = 51.1825m – 50.0975m
S = 1.085m
1.085m -------- 220m
Xm -------- 100%
X =

X = 0.4932%

6. Condiciones de la Razante:

Cota = 50.50m
Pendiente = 0.02%

Para 20mDistanciaAcumulada | Razante(m) | CotaRazante |
0 | 0 | 50.50m |
20 | 0.004 | 50.496m |
40 | 0.008 | 50.492m |
60 | 0.012 | 50.488m |
80 | 0.016 | 50.484m |
100 | 0.020 | 50.48m |
120 | 0.024 | 50.476m |
140 | 0.028 | 50.472m |
100m ----- 0.02m
20 ----- X X = 0.004m
Para 40m

100m ----- 0.02m
40 ----- X X = 0.008m

Para 60m

100m ----- 0.02m
60 ----- X X =0.012m

Para 80m

100m ----- 0.02m
80 ----- X X = 0.016m
160 | 0.032 | 50.468m |
180 | 0.036 | 50.464m |
200 | 0.040 | 50.46m |
220 | 0.044 | 50.456m |

Para 100m

100m ----- 0.02m
100 ----- X X = 0.020m

Para 120m

100m ----- 0.02m
120 ----- X X = 0.024m

Para 140m

100m ----- 0.02m
135.15----- X X = 0.028m

Para 160m

100m ----- 0.02m
160----- X X = 0.032m

Para 180m

100m ----- 0.02m
180 ----- X X = 0.036m

Para 200m

100m ----- 0.02m
200 ----- X X = 0.040m

Para 220m

100m ----- 0.02m
220 ----- X X = 0.044m

7. Determinación de alturas de corte y relleno :

H =│ Cota Razante-Cota Terreno │

H (0) =│50.0975m-50.50m│= 0.4025m

H (20) =│50.3725m-50.496m│= 0.1235m

H (40) =│50.8705m-50.492m│=...