deber inderterminacion estatica interna
DESCRIPTION
DETERMINACION ESTATICA INTERNA EN ARMADURATRANSCRIPT
UNIVERSIDAD CENTRAL DEL ECUADOR FACULTAD DE INGENIERÍA CIENCIAS FÍSICAS Y MATEMÁTICAS
2015
SONY CARRERA DE INGENIERÍA CIVIL
17/10/2015
ESTRUCTURAS II
NOMBRE: GREFA AGUINDA JUAN EUCLIDES
CURSO: SEXTO
PARALELO: 1
TEMA: INDETERMINACIÓN ESTÁTICA INTERNA
DE ARMADURA PLANA
FECHA DE ENTREGA: 19 / 10 / 2015
SEMESTRE SEPTIEMBRE 2014- FEBRERO 2015
Resolver el siguiente problema con indeterminación estática interna (IEI).
3 3
4
5
5
5 5
A1 A1
A1 A1
A2A2
d11
1
1
5
1
d11
d12
d12
A2
d22
d22
d21
d21
A21
[N]
[N]
[n]
[n]
A
B F C
ED
A
B F C
ED A
B F C
ED A
B F C
ED
A
B F C
ED A
B F C
ED A
B F C
ED
RAz RBz
RAz RBz
Con N = (1)*A1
∑ 𝑀𝐴𝑛𝑖=1 = 0
𝑅𝐵𝑧 ∗ 6 + 1 ∗3
5∗ 4 − 1 ∗
4
5∗ 3 = 0
𝑹𝑩𝒛 = 𝟎
∑ 𝐹𝑧 = 0
𝑅𝐴𝑧 + 𝑅𝐵𝑧 + 14
5− 1
4
5= 0
𝑹𝑨𝒛 = 𝟎
∑ 𝐹𝑍 𝐴 = 0
−𝐴𝐵 +4
5= 0 𝑨𝑩 =
𝟒
𝟓 𝑪
∑ 𝐹𝑋 𝐴 = 0
−𝐴𝐸 +3
5= 0 𝑨𝑬 =
𝟑
𝟓 𝑪
∑ 𝐹𝑍 𝐵 = 0
4
5− 𝐵𝐸 ∗
4
5= 0 𝑩𝑬 = 𝟏 𝑻
∑ 𝐹𝑋 𝐵 = 0
−𝐵𝐹 +3
5= 0 𝑩𝑭 =
𝟑
𝟓 𝑪
∑ 𝐹𝑍 𝐹 = 0
𝐸𝐹 −4
5= 0 𝑬𝑭 =
𝟒
𝟓 𝑪
∑ 𝐹𝑍 𝐹 = 0
𝐵𝐹 −3
5+ 𝐹𝐶 = 0 𝑭𝑪 = 𝟎
𝑫𝑪 = 𝑫𝑬 = 𝑪𝑬 = 𝟎
BARRA N*A1 n1 L N.n.L
1 - 3/5 - 3/5 3 1 2/25
2 0 0 3 0
3 - 3/5 - 3/5 3 1 2/25
4 0 0 3 0
5 - 4/5 - 4/5 4 2 14/25
6 1 1 5 5
7 - 4/5 - 4/5 4 2 14/25
8 0 0 5 0
9 0 0 4 0
∑ =d 11= 12 7/25 A1
AE
CON N =1*A2 (ARMADURA SIMÉTRICA)
BARRA N*A2 n2 L N.n.L
1 0 0 3 0
2 - 3/5 - 3/5 3 1 2/25
3 0 0 3 0
4 - 3/5 - 3/5 3 1 2/25
5 0 0 4 0
6 0 0 5 0
7 - 4/5 - 4/5 4 2 14/25
8 1 1 5 5
9 - 4/5 - 4/5 4 2 14/25
∑ =d 22= 12 7/25 A2
AE
BARRA N*A2 n1 L N.n.L
1 0 - 3/5 3 0
2 - 3/5 0 3 0
3 0 - 3/5 3 0
4 - 3/5 0 3 0
5 0 - 4/5 4 0
6 0 1 5 0
7 - 4/5 - 4/5 4 2 14/25
8 1 0 5 0
9 - 4/5 0 4 0
∑ =d 12= 2 14/25 A2
AE
BARRA N*A1 n2 L N.n.L
1 - 3/5 0 3 0
2 0 - 3/5 3 0
3 - 3/5 0 3 0
4 0 - 3/5 3 0
5 - 4/5 0 4 0
6 1 0 5 0
7 - 4/5 - 4/5 4 2 14/25
8 0 1 5 0
9 0 - 4/5 4 0
∑ =d 21= 2 14/25 A1
AE
5
A
B F C
ED
5/2 5/2
-5/225/8 0
25/8-5/2
0 0
-15/8 -15/8
BARRA Nq n1 L N.n.L
1 0 - 3/5 3 0
2 0 0 3 0
3 -1 7/8 - 3/5 3 3 3/8
4 -1 7/8 0 3 0
5 -2 1/2 - 4/5 4 8
6 3 1/8 1 5 15 5/8
7 0 - 4/5 4 0
8 3 1/8 0 5 0
9 -2 1/2 0 4 0
∑ =d 1q= 27
AE
BARRA Nq n2 L N.n.L
1 0 0 3 0
2 0 - 3/5 3 0
3 -1 7/8 0 3 0
4 -1 7/8 - 3/5 3 3 3/8
5 -2 1/2 0 4 0
6 3 1/8 0 5 0
7 0 - 4/5 4 0
8 3 1/8 1 5 15 5/8
9 -2 1/2 - 4/5 4 8
∑ =d 2q= 27
AE
DESPLAZAMIENTOS
𝑑1 =𝐿
𝐴𝐸∗ 𝐴1 =
5
𝐴𝐸𝐴1 𝑑2 =
𝐿
𝐴𝐸𝐴2 𝒅𝟏 =
𝟓
𝑨𝑬𝑨𝟏 𝒅𝟐 =
𝟓
𝑨𝑬𝑨𝟐
1) 𝑑1𝑞 + 𝑑11 + 𝑑12 + 𝑑1 = 0
2) 𝑑2𝑞 + 𝑑21 + 𝑑22 + 𝑑2 = 0
27 +307
25𝐴1 +
64
25𝐴2 + 5𝐴1 = 0
27 +64
25𝐴1 +
307
25𝐴2 + 5𝐴2 = 0
27 + (307
25+ 5) 𝐴1 +
64
25𝐴2 = 0
27 +64
25𝐴1 + (
307
25+ 5) 𝐴2 = 0
`𝑨𝟏 = −𝟏. 𝟑𝟔𝟏 = 𝑨𝑭
𝑨𝟐 = −𝟏. 𝟑𝟔𝟏 = 𝑭𝑫
5
A
B F C
ED
5/2 5/2
-1.361
1.76
4
2.1
78
1.764
-1.3
61
-1.4
11
-1.058-1.058-1
.411.
0.817 0.817
∑ 𝐹𝐴𝑥 = 0
𝐴𝐸 − 1.361 ∗3
5+ 5 = 0
𝑨𝑬 = 𝟎. 𝟖𝟏𝟕 T
∑ 𝐹𝐴𝑧 = 0
−𝐴𝐵 − 1.361 ∗4
5+ 5 = 0
𝑨𝑩 = 𝟏. 𝟒𝟏𝟏 𝑪
∑ 𝐹𝐵𝑧 = 0
1.411 − 𝐵𝐸 ∗4
5= 0
𝑩𝑬 = 𝟏. 𝟕𝟔𝟒 𝑻
∑ 𝐹𝐵𝑥 = 0
−𝐵𝐹 + 1.764 ∗3
5= 0
𝑩𝑭 = 𝟏. 𝟎𝟓𝟖 𝑪
∑ 𝐹𝐹𝑧 = 0
−𝐹𝐸 + 1.361 ∗4
5+ 1.361 ∗
4
5= 0
𝑭𝑬 = 𝟐. 𝟏𝟕𝟖 𝑻
𝑃𝑂𝑅 𝑆𝐼𝑀𝐸𝑇𝑅Í𝐴 𝐷𝐸 𝐿𝐴 𝐴𝑅𝑀𝐴𝐷𝑈𝑅𝐴
𝑬𝑫 = 𝑨𝑬 = 𝟎. 𝟖𝟏𝟕 𝑻
𝑫𝑪 = 𝑨𝑩 = 𝟏. 𝟒𝟏𝟏 𝑪
𝑩𝑬 = 𝑬𝑪 = 𝟏. 𝟕𝟔𝟒 𝑻
𝑩𝑭 = 𝑭𝑪 = 𝟏. 𝟎𝟓𝟖 𝑪
DIAGRAMA DE FUERZAS AXIALES
-1.361
1.76
4
2.17
8
1.764
-1.3
61
-1.4
11
-1.058-1.058
-1.4
11.
0.817 0.817
[N]