pavimentación
Sede regional, Estelí
Ingeniería Civil
Estática -
Reacciones
Docente: Ing. Sergio Navarro H
Realice la siguiente reacción de:
2 KN/m
5 KN
2KN/m
4m
A
3m
3m
(2)(3)=6
5 KN
5
4
3
4m
MA
5/2
2
A
RAX
RAY
3m
3
3/2
5
4
α
3
Determine las reacciones en los apoyos
Carga:
X=(20KN/m)(Sen 50º)=15.3208 KN/m
Y=(20KN/m)(Cos 50º)= 12.8557 KN/m
F1x=
=
= 38.302KN
=
F1y=
=
= 32.13925KN
=
F2= b.h = (5 m)(20KN/m)= 100KN
Carga:
= b/2= 5 m /2 = 2.5 m
X=15.3208 KN/m
Y=12.8557 KN/m
F3x=
=
= 38.302KN
=
F3y=
=
= 32.13925KN
=
#1.
RAX + 38.302 KN - 38.302KN = 0
RAX = 0
#2.
RAY – 32.13925 KN – 100 – 32.13925KN= 0
RAY= 164.2785 KN
#3.-M-(38.302KN*
)–(32.13925*
)–(100KN*7.5 m)+(38.302KN*
M= -1,232.08875KN.m ò M= 1232.08875Kn.m
) –(32.13925KN)(
300N/m
100N/m
RAX
50N/m
210N.m
RBY
RAY
2m
4m
2m
200N.m)=0
Determine las reacciones en los apoyos:
a) Fuerzas puntuales:
1.
2.
RBY
RAX
RAY
23 N
50 N
RAX
RAY
RBY
Determine las reacciones en el punto A
+
3108.1831Ejercicio Nº 16
+
+
+
Sustituimos
1)
2)
en (1)
3)
(1)
∑ Fx = 0
+
80-10 + RAX = 0
RAX = - 70 N
∑ FY= 0
RAX = 70 N
+
70 + RAX + RBY = 0
RAX + RBY = 70(1)
∑ MA= 0
+
-80 (8/3) – 70 (7/2) + 10 (6) + RBY (7) = 0
- 398.33 + RBY (7) =0
RBY= 398.33 N.m
7m
RBY = 56.9 N
Sustituyo en (1)
RAY + RBY = 70
RAY + 56.9 N = 70
RAY = (70-56.9)N
RAY = 13.1 N
+
3( ) + RBX + (5) (7) = 0
RBX= -37.12N
3
2
2
2
A
+
5kn/m
-3( ) -2 -2 -2 -2 + RAY + RBY = 0
7mts
RAY + RBY = 10.12N
B
+
( ) (7) + 3( ) (6) + 2(4)+2(2) -2(2) -5(7) (3.5) + RAY (6)
RAY = +116.62
6
RAY = 19.44
RBY = (10.12 – 19.44) N
RBY = -9.32
50N/M
25N/M
20N/M
RAx
RAy
A
2m
B
5m
F1 25N F2 75N F3 100N
0.6m...
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