Metodo de Rigidez
ANALISIS ESTRUCTURAL II PROBLEMAS POR EL MÉTODO DE RIGIDEZ Problema 1 T2 ø 12 T/m K2 30 T-m K3 T1 20 T T 3 4.0 K1
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ANALISIS ESTRUCTURAL II
PROBLEMAS POR EL MÉTODO DE RIGIDEZ Problema 1
T2 ø
12 T/m
K2 30 T-m
K3
T1 20 T
T 3 4.0 K1 4.0
2.0
2.0
DATOS: 𝐸 = 2 ∗ 106 𝑇/𝑚2
𝑏 ∗ ℎ = 0.3 ∗ 0.3 𝑚2
𝐾1 = 1000 𝑇/𝑚
𝐾2 = 2000 𝑇/𝑚
𝐾3 = 1000 𝑇 − 𝑚/𝑟𝑎𝑑
𝛼 = 10−5 °𝐶 −1
∆𝑇1 = 1000 °𝐶
∆𝑇2 = 2000 °𝐶
∆= 0.01 𝑚
∅ = 0.02 𝑟𝑎𝑑
∆𝑇3 = 3000 °𝐶
4√2 cm A
3√2
√2
B
Hallar: −𝐷𝑀𝐹 −𝑀𝐴 𝑅𝑉𝐷 SOLUCION 1) Definimos Q – D Y q – d 2 6
5
3
1
1
2
1 2
2
1 3
1
Q-D
4
q-d
1
4
4
AGUEDO TORRES ALEXANDER
2
3 1
5
6
UNASAM-FIC
ANALISIS ESTRUCTURAL II
2) Resolvemos el problema primario.
Para cargas externas.
RL 2 12 T/m
12 T/m
30 T-m N1
RL1 B N2 4 4 12 12
4 12
32
C
4 12
4
4
12
32 D
12
RL
10
2
14
20 T
2
E
A
10 AR1
RL 4
10 AR2
NUDO B B
N2
N1 20 + 12
∑ 𝑭𝒀 = 𝟎 𝑵𝟏 = −𝟑𝟐√𝟐 ∑ 𝑭𝑿 = 𝟎 𝑵𝟐 = −𝟑𝟐
−34 26 𝑅𝐿 = { } −32 0
0 𝐴𝑅𝐿 = { } 10
AGUEDO TORRES ALEXANDER
𝐴𝑀𝐿
0 0 −4 0 = 0 0 0 0 {0}
UNASAM-FIC
ANALISIS ESTRUCTURAL II
Por temperatura
R T2 B 45
R T1 90 45
90
90
90
135
90
135 D 22.5
C 22.5 22.5
45
R T3
22.5 135
45
135
E
A
22.5 135
45 AP1
R T4
22.5 AP 2
45 22.5 𝑅𝑇 = { } 0 0
−45 𝐴𝑅𝑇 = { } −22.5
𝐴𝑀𝑇
45 −45 90 −135 = −135 135 0 0 { 0 }
Por deformación previa.
RP2 B N1 6.39
R P1 N2
C
9.56
9.56 D
RP
6.76 6.76
31.83 6.76 E
A 31.83
RP4
AR1 AR2
AGUEDO TORRES ALEXANDER
UNASAM-FIC
ANALISIS ESTRUCTURAL II
NUDO B
B
N2
N1 6.76 ∑ 𝑭𝒀 = 𝟎 𝒄𝒐𝒔(𝟒𝟓) ∗ 𝑵𝟏 = −𝟔. 𝟕𝟔 ∗ 𝒄𝒐𝒔(𝟒𝟓) 𝑵𝟏 = −𝟔. 𝟕𝟔 ∑ 𝑭𝑿 = 𝟎 𝑵𝟐 = −𝟔. 𝟕𝟔 ∗ √𝟐 𝑵𝟐 = 𝟗. 𝟓𝟔
6.39 0 𝑅𝑃 = { } −9.56 0
−31.83 𝐴𝑅𝑃 = { } 0
𝐴𝑀𝑃
31.83 6.39 0 0 = 0 0 0 0 { 0 }
Por asentamiento de apoyo. R R2
R R1
0.02
R R3
R R4
AGUEDO TORRES ALEXANDER
UNASAM-FIC
ANALISIS ESTRUCTURAL II
60
R R2
20 20.25 R R1 B N2 N1
20.25
60 20.25
C 20.25
20.25
40.13
60
D R R3 40.13
40.13
20.25 20.25 40.13 20.25
E
A
40.13 AR1
60
R R4
40.13 AR2
20.25 60.38 𝑅𝑅 = { } 40.25 0
0 𝐴𝑅𝑅 = { } −40.13
𝐴𝑀𝑅
0 0 20.25 −60 = 0 0 0 20 { 60 }
𝑹 = 𝑹𝑳 + 𝑹𝑻 + 𝑹𝑷 + 𝑹𝑹 𝑸 = −𝑹
−34 6.39 45 20.25 26 0 22.5 60.38 𝑅={ }+ { }+{ }+{ } −32 −9.56 0 40.25 0 0 0 0 37.64 10.88 𝑅={ } −1.31 0
AGUEDO TORRES ALEXANDER
UNASAM-FIC
ANALISIS ESTRUCTURAL II
−37.64 −108.88 𝑄={ } 1.31 0 3) Hallamos 𝑨: 𝒅 = 𝑨 𝑫
𝑫𝟏 = 𝟏
𝑫𝟑 = 𝟏
𝑫𝟐 = 𝑫𝟑 = 𝑫𝟒 = 𝟎
𝑫𝟐 = 𝟏
𝑫𝟏 = 𝑫𝟐 = 𝑫𝟒 = 𝟎
𝑨=
0 1 1 0 0 0 0 0 0
AGUEDO TORRES ALEXANDER
𝑫𝟒 = 𝟏
0 0 0 0.5 0 0.5 0 0 0.5
-0.25 -0.25 0.25 0.25 -0.25 -0.25 0 1 0
𝑫𝟏 = 𝑫𝟑 = 𝑫𝟒 = 𝟎
𝑫𝟏 = 𝑫𝟐 = 𝑫𝟑 = 𝟎
0 0 0 0.25 0.25 0 1 0 0
UNASAM-FIC
ANALISIS ESTRUCTURAL II
4) Hallamos 𝑲𝑬 : 2
𝑲𝟏 =
𝟐𝑬𝑰 𝟐 [ 𝑳 𝟏
𝑲𝟏 = [
𝟏 ] 𝟐
; 𝑬𝑰 = 𝟏𝟑𝟓𝟎 ; 𝑳 = 𝟒√𝟐
𝟗𝟓𝟒. 𝟓𝟗 𝟒𝟕𝟕. 𝟑𝟎 ] 𝟒𝟕𝟕. 𝟑𝟎 𝟗𝟓𝟒. 𝟓𝟗
1 2
𝟐 𝟒 𝑲𝟐 = 𝑬𝑰 [ ] 𝟒 𝟏𝟒
𝑲𝟐 = [
𝑲𝟑 =
𝟐𝟕𝟎𝟎 𝟓𝟒𝟎𝟎 ] 𝟓𝟒𝟎𝟎 𝟏𝟖𝟗𝟎𝟎
𝟐𝑬𝑰 𝟐 [ 𝑳 𝟏
𝑲𝟑 = [
; 𝑬𝑰 = 𝟏𝟑𝟓𝟎 ;
𝟏 ] 𝟐
; 𝑬𝑰 = 𝟏𝟑𝟓𝟎 ; 𝑳 = 𝟒
𝟏𝟑𝟓𝟎 𝟔𝟕𝟓 ] 𝟔𝟕𝟓 𝟏𝟑𝟓𝟎
𝑲𝟒 = [𝟏𝟎𝟎𝟎] 𝑲𝟓 = [𝟐𝟎𝟎𝟎] 𝑲𝟓 = [𝟑𝟎𝟎𝟎]
AGUEDO TORRES ALEXANDER
UNASAM-FIC
ANALISIS ESTRUCTURAL II
954.59 477.30 0 0 0 0 0 0 0
𝑲𝑬 =
477.30 954.59 0 0 0 0 0 0 0
0 0 2700 5400 0 0 0 0 0
0 0 5400 18900 0 0 0 0 0
0 0 0 0 1350 675 0 0 0
0 0 0 0 675 1350 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 1000 0 0 2000 0 0
0 0 0 0 0 0 0 0 3000
5) Hallamos 𝑲: 𝑲 = 𝑨𝑻 𝑲𝑬 𝑨 954.59
477.30
0
0
0
0
0
0
0
0
0
-0.25
477.30
954.59
0
0
0
0
0
0
0
1
0
-0.25
0 0
0
1
1
0
0
0
0
0
0
0
0
2700
5400
0
0
0
0
0
1
0
0.25
0
0
0
0
0.5
0
0.5
0
0
0.5
0
0
5400
18900
0
0
0
0
0
0
0.5
0.25
0.25
-0.25
-0.25
0.25
0.25
-0.25
-0.25
0
1
0
0
0
0
0
1350
675
0
0
0
0
0
-0.25
0.25
0
0
0
0.25
0.25
0
1
0
0
0
0
0
0
675
1350
0
0
0
0
0.5
-0.25
0
0
0
0
0
0
0
1000
0
0
0
0
0
1
0
0
0
0
0
0
0
2000
0
0
0
1
0
0
0
0
0
0
0
0
0
3000
0
0.5
0
0
𝑲=
𝑲=
3654.59 2700.00 1667.03 1350.00
2700.00 5812.50 2784.38 2446.88
1667.03 2784.38 4457.11 1392.19
1350.00 2446.88 1392.19 2265.63
6) Hallamos 𝑫: 𝑸=𝑲𝑫 𝑫 = 𝑲−𝟏 𝑸
𝑲−𝟏
0.000429734 -0.000149479 = -4.67668E-05 -6.58873E-05
-0.00014948 0.000420401 -0.00011475 -0.00029446
AGUEDO TORRES ALEXANDER
-4.67668E-05 -6.5887E-05 -0.000114745 -0.00029446 0.000329332 -5.0577E-05 -5.05773E-05 0.00082973
UNASAM-FIC
ANALISIS ESTRUCTURAL II
0.000429734 -0.000149479 𝑫= -4.67668E-05 -6.58873E-05
-0.00014948 0.000420401 -0.00011475 -0.00029446
-6.5887E-05 -0.00029446 -5.0577E-05 0.00082973
-4.67668E-05 -0.000114745 0.000329332 -5.05773E-05
-37.64 -108.88 1.31 0
0 -0.040 0.015 0.034
𝑫=
7) Hallamos 𝑨𝑴 : 𝑨𝑴 = 𝒒𝒑𝒓𝒊𝒎𝒂𝒓𝒊𝒐 + 𝒒𝒄𝒐𝒎𝒑𝒍𝒆𝒎𝒆𝒏𝒕𝒂𝒓𝒊𝒐 𝑨𝑴 = 𝑨𝑴𝑳 + 𝑨𝑴𝑻 + 𝑨𝑴𝑷 + 𝑨𝑴𝑹 + 𝑲𝑬 ∗ 𝒅 𝑨𝑴 = 𝑨𝑴𝑳 + 𝑨𝑴𝑻 + 𝑨𝑴𝑷 + 𝑨𝑴𝑹 + 𝑲𝑬 ∗ 𝑨 ∗ 𝑫 𝒒𝒑𝒓𝒊𝒎𝒂𝒓𝒊𝒐 = 𝑨𝑴𝑳 + 𝑨𝑴𝑻 + 𝑨𝑴𝑷 + 𝑨𝑴𝑹 𝒒𝒄𝒐𝒎𝒑𝒍𝒆𝒎𝒆𝒏𝒕𝒂𝒓𝒊𝒐 = 𝑲𝑬 ∗ 𝑨 ∗ 𝑫
𝒒𝒑𝒓𝒊𝒎𝒂𝒓𝒊𝒐 =
𝒒𝒄𝒐𝒎𝒑𝒍𝒆𝒎𝒆𝒏𝒕𝒂𝒓𝒊𝒐 =
0 0 -4 0 0 0 0 0 0
+
45 -45 90 -135 -135 135 0 0 0
31.83 6.39 0 0 0 0 0 0 0
+
0 0 20.25 -60 0 0 0 20 60
+
=
954.59
477.30
0
0
0
0
0
0
0
0
477.30
954.59
0
0
0
0
0
0
0
0
0
2700
5400
0
0
0
0
0
0
5400
18900
0
0
0
0
0
0
0
1350
675
0
0
0
0
675
0
0
0
0
0
0
0
0
0
0
0
0
0
AGUEDO TORRES ALEXANDER
76.83 -38.61 106.25 -195 -135 135 0 20 60 0
0
-0.25
1
0
-0.25
0
0
1
0
0.25
0
4E-05
0
0
0
0.5
0.25
0.25
-0.04
0
0
0
0
0
-0.25
0.25
0.015
1350
0
0
0
0
0.5
-0.25
0
0.034
0
1000
0
0
0
0
0
1
0
0
0
2000
0
0
0
1
0
0
0
0
0
3000
0
0.5
0
0
UNASAM-FIC
ANALISIS ESTRUCTURAL II
𝒒𝒄𝒐𝒎𝒑𝒍𝒆𝒎𝒆𝒏𝒕𝒂𝒓𝒊𝒐 =
-5.24 -5.22 -32.42 -128.50 -9.40 -28.82 34.47 29.37 -60.45
𝑨𝑴 = 𝒒𝒑𝒓𝒊𝒎𝒂𝒓𝒊𝒐 + 𝒒𝒄𝒐𝒎𝒑𝒍𝒆𝒎𝒆𝒏𝒕𝒂𝒓𝒊𝒐
𝑨𝑴 =
76.83 -38.61 106.25 -195 -135 135 0 20 60
+
𝑨𝑴 =
AGUEDO TORRES ALEXANDER
-5.24 -5.22 -32.42 -128.50 -9.40 -28.82 34.47 29.37 -60.45
=
71.59 -43.83 73.83 -323.50 -144.40 106.18 34.47 49.37 -0.45
71.59 -43.83 73.83 -323.50 -144.40 106.18 34.47 49.37 -0.45
UNASAM-FIC
ANALISIS ESTRUCTURAL II
Problema 02:
K2
2.0
9 T/m
Rotula B
A
E
K1
T 20 T
5
m T/
T C
D 4
2
4
ø
(DE) Barra tipo armadura 𝐸 = 2 ∗ 105 𝑇/𝑚2
𝐴 = 10 𝑐𝑚2
DATOS: 𝐸 = 2 ∗ 106 𝑇/𝑚2
𝑏 ∗ ℎ = 0.3 ∗ 0.3 𝑚2
𝐾1 = 1000 𝑇/𝑚
𝐾2 = 2000 𝑇 − 𝑚/𝑟𝑎𝑑
𝛼 = 10−5 °𝐶 −1
∆𝑇 = 1000 °𝐶
∆= 0.01 𝑚
∅ = 0.02 𝑟𝑎𝑑
0.075 ∗ √2 cm
C 1.5√2
2.5√2
Hallar: −𝐷𝑀𝐹 −𝑀𝐴 𝑀𝐶
AGUEDO TORRES ALEXANDER
UNASAM-FIC
ANALISIS ESTRUCTURAL II
SOLUCION 1)
Definimos Q – D Y q – d 5 1 1
A
B 1
1
E 4
2
2
2
1
1
2
2
3
1 1 1
3
Q-D
C
D
4
q-d
2)
Resolvemos el problema primario. Para cargas externas. RL1 9 T/m
9 T/m A AR1
9
3
3
9
9
B 3 10 2
5
E
3 9
RL 2 20 T
m T/
RL 3
10 2
C
D
AR
𝟗 𝑹𝑳 = {𝟐𝟖} 𝟎
𝟑 𝑨𝑹𝑳 = { } 𝟎
𝑨𝑴𝑳
−𝟑 𝟐𝟖 𝟎 = 𝟎 −𝟏𝟎√𝟐 𝟎 { 𝟎 }
AGUEDO TORRES ALEXANDER
UNASAM-FIC
ANALISIS ESTRUCTURAL II
Por temperatura R T1 B
A 45
45
AR1 45
E
45 2
R T2
2 R T3 2 45
C
D
AR2 45
BARRA TIPO ARMADURA 𝑃𝐿
𝛼∆𝑇𝐿 = 𝐸𝐴
𝑃 = 𝐸𝐴𝛼∆𝑇
𝑃 = 2 ∗ 105 ∗ 10−3 ∗ 10−5 ∗ 103 𝑃 =2𝑇 𝑅𝑇2 = 45 − 2 ∗ √2 = 42.17 BARRA 1 Y 2 Barra doblemente empotrada 𝑀1 = 𝑀2 = 𝑀=
𝛼𝐸𝐼∆𝑇 ℎ
10−5 ∗ 1350 ∗ 1000 = 45 0.3
0 𝑅𝑇 = {42.17} 45
45 𝐴𝑅𝑇 = { } −45
AGUEDO TORRES ALEXANDER
𝐴𝑀𝑇
−45 42.17 45 = 45 −2 0 { 0 }
UNASAM-FIC
ANALISIS ESTRUCTURAL II
Por deformación previa.
R P1 B
A
E RP2
AR1
33.41
R P3 4.78
C
D
AR 4.78
PARA LA BARRA 2
23.86
47.73
28.64
14.32
𝑴′𝟏 =
𝟒𝑬𝑰 𝟒 ∗ 𝟏𝟑𝟓𝟎 ∗∅ = ∗ 𝟎. 𝟎𝟓 = 𝟒𝟕. 𝟕𝟑 𝑳 𝟒√𝟐
𝑴′𝟐 =
𝟐𝑬𝑰 𝟐 ∗ 𝟏𝟑𝟓𝟎 ∗∅ = ∗ 𝟎. 𝟎𝟓 = 𝟐𝟑. 𝟖𝟔 𝑳 𝟒√𝟐
𝑴′′ 𝟏 =
𝟒𝑬𝑰 𝟐 ∗ 𝟏𝟑𝟓𝟎 ∗∅= ∗ 𝟎. 𝟎𝟑 = 𝟏𝟒. 𝟑𝟐 𝑳 𝟒√𝟐
𝑴′′ 𝟐 =
𝟐𝑬𝑰 𝟒 ∗ 𝟏𝟑𝟓𝟎 ∗∅= ∗ 𝟎. 𝟎𝟑 = 𝟐𝟖. 𝟔𝟒 𝑳 𝟒√𝟐
0 𝑅𝑃 = { 0 } −33.41
0 𝐴𝑅𝑃 = { } 4.78
AGUEDO TORRES ALEXANDER
𝐴𝑀𝑃
0 0 33.41 = −4.78 0 0 { 0 }
UNASAM-FIC
ANALISIS ESTRUCTURAL II
Por asentamiento de apoyo. R R1 B
A
E R R2
=0.01
R R3
C
D
ø
R R1 B
A
E R R2
10
AR1
9.55 R R3 19.09
C
D
AR2
10 𝑅𝑅 = { 0 } 9.55
0 𝐴𝑅𝑅 = { } 19.09
𝐴𝑀𝑅
0 0 −9.55 = −19.09 0 0 { 0 }
𝑹 = 𝑹𝑳 + 𝑹𝑻 + 𝑹𝑷 + 𝑹𝑹 𝑸 = −𝑹
AGUEDO TORRES ALEXANDER
UNASAM-FIC
ANALISIS ESTRUCTURAL II
𝟗 𝟎 𝟎 𝟏𝟎 𝑹 = {𝟐𝟖} + {𝟒𝟐. 𝟏𝟕} + { 𝟎 } + { 𝟎 } 𝟎 𝟒𝟓 −𝟑𝟑. 𝟒𝟏 𝟗. 𝟓𝟓 𝟏𝟗 𝑹 = {𝟕𝟎. 𝟏𝟕} 𝟐𝟏. 𝟏𝟒 −𝟏𝟗 𝑸 = {− 𝟕𝟎. 𝟏𝟕 } −𝟐𝟏. 𝟏𝟒
3)
Hallamos 𝑨: 𝒅 = 𝑨 𝑫 𝑫𝟏 = 𝟏
𝑫𝟐 = 𝑫𝟑 = 𝟎
𝑫𝟐 = 𝟏
𝑫𝟏 = 𝑫𝟑 = 𝟎
AGUEDO TORRES ALEXANDER
UNASAM-FIC
ANALISIS ESTRUCTURAL II
𝑫𝟑 = 𝟏
-0.25 0 -0.25 1 0 0 0 0 A= 0 1.41 1 0 0 1 4)
𝑫𝟏 = 𝑫𝟐 = 𝟎
0 0 -1 0 0 0 1
Hallamos 𝑲𝑬 : 1 2
AGUEDO TORRES ALEXANDER
UNASAM-FIC
ANALISIS ESTRUCTURAL II 𝟐 𝟒 𝑲𝟏 = 𝑬𝑰 [ ] 𝟒 𝟏𝟒 𝑲𝟏 = [
; 𝑬𝑰 = 𝟏𝟑𝟓𝟎
𝟐𝟕𝟎𝟎 𝟓𝟒𝟎𝟎 ] 𝟓𝟒𝟎𝟎 𝟏𝟖𝟗𝟎𝟎 2
𝑲𝟐 =
𝟐𝑬𝑰 𝟐 [ 𝑳 𝟏
𝑲𝟐 = [
𝟏 ] 𝟐
; 𝑬𝑰 = 𝟏𝟑𝟓𝟎 ; 𝑳 = 𝟒√𝟐
𝟗𝟓𝟒. 𝟓𝟗 𝟒𝟕𝟕. 𝟑𝟎 ] 𝟒𝟕𝟕. 𝟑𝟎 𝟗𝟓𝟒. 𝟓𝟗 𝑬𝑨
𝑲𝟑 = [ ] 𝑳
; 𝑬𝑨 = 𝟐𝟎𝟎 ; 𝑳 = 𝟒√𝟐
𝑲𝟑 = [𝟐𝟓√𝟐] 𝑲𝟒 = [𝟏𝟎𝟎𝟎] 𝑲𝟓 = [𝟐𝟎𝟎𝟎] 2700 5400 0 0 0 0 0
KE =
5)
5400 18900 0 0 0 0 0
0 0 954.59 477.30 0 0 0
0 0 477.30 954.59 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 35.36 0 0 0 1000 0 0 0 2000
Hallamos 𝑲:
𝑲 = 𝑨𝑻 𝑲𝑬 𝑨
K=
2700
5400
0
0
0
0
0
-0.25
0
0
5400
18900
0
0
0
0
0
-0.25
1
0
0
0
0
1
0
0
0
954.59
477.30
0
0
0
0
0
-1
1
0
0
1.41
0
1
0
0
477.30
954.59
0
0
0
0
0
0
0
-1
0
0
0
1
0
0
0
0
35.36
0
0
0
1.41
0
0
0
0
0
0
1000
0
1
0
0
0
0
0
0
0
0
2000
0
1
1
-0.25
-0.25
0 0
K=
3025 -6075 0
AGUEDO TORRES ALEXANDER
-6075 20970.71 2000
0 2000 2954.59
UNASAM-FIC
ANALISIS ESTRUCTURAL II
6) Hallamos 𝑫: 𝑸=𝑲𝑫 𝑫 = 𝑲−𝟏 𝑸
K¯ˡ=
0.00087 0.00027 -0.00018
0.00027 -0.00018 0.00013 -0.00009 -0.00009 0.00040
D=
0.00087 0.00027 -0.00018
0.00027 -0.00018 0.00013 -0.00009 -0.00009 0.00040
D=
-19 -70.17 -21.14
-0.0317 -0.0127 0.0014
7) Hallamos 𝑨𝑴 : 𝑨𝑴 = 𝒒𝒑𝒓𝒊𝒎𝒂𝒓𝒊𝒐 + 𝒒𝒄𝒐𝒎𝒑𝒍𝒆𝒎𝒆𝒏𝒕𝒂𝒓𝒊𝒐 𝑨𝑴 = 𝑨𝑴𝑳 + 𝑨𝑴𝑻 + 𝑨𝑴𝑷 + 𝑨𝑴𝑹 + 𝑲𝑬 ∗ 𝒅 𝑨𝑴 = 𝑨𝑴𝑳 + 𝑨𝑴𝑻 + 𝑨𝑴𝑷 + 𝑨𝑴𝑹 + 𝑲𝑬 ∗ 𝑨 ∗ 𝑫 𝒒𝒑𝒓𝒊𝒎𝒂𝒓𝒊𝒐 = 𝑨𝑴𝑳 + 𝑨𝑴𝑻 + 𝑨𝑴𝑷 + 𝑨𝑴𝑹 -3 28 0
qprimario=
-45 42.17 45
0
+
45
-14.14 0 0
qcomp.=
0 0 33.41 +
-2 0 0
-4.78
0 0 -9.55 +
0 0 0
-19.09 0 0 0
-48.00 70.17 68.86 =
21.13 -16.14 0.00 0.00
2700 5400 0
5400 18900 0
0 0 0 0 954.59 477.30
0 0 0
0 0 0
0 0 0
-0.25 -0.25 0
0 1 0
0 0 -1
-0.0317
0
0
477.30 954.59
0
0
0
0
0
0
-0.0127
0 0 0
0 0 0
0 1 0
1.41 0 1
0 0 1
0.0014
0 0 0
0 0 0
AGUEDO TORRES ALEXANDER
35.36 0 0 0 1000 0 0 0 2000
UNASAM-FIC
ANALISIS ESTRUCTURAL II
-4.182 -46.773 -1.361 qcomp.=
-0.681 -0.634 -31.739 -22.501
𝑨𝑴 = 𝒒𝒑𝒓𝒊𝒎𝒂𝒓𝒊𝒐 + 𝒒𝒄𝒐𝒎𝒑𝒍𝒆𝒎𝒆𝒏𝒕𝒂𝒓𝒊𝒐 -52.182 23.397 67.499 AM=
20.449 -16.776 -31.739 -22.501
AGUEDO TORRES ALEXANDER
UNASAM-FIC
ANALISIS ESTRUCTURAL II
8) Hallamos 𝑨𝑹 : 𝐴𝑅 = 𝑟𝑃 + 𝑟𝐶 𝐴𝑅 = 𝐴𝑅𝐿 + 𝐴𝑅𝑇 + 𝐴𝑅𝑃 + 𝐴𝑅𝑅 + 𝐴𝑅𝐷 𝐷 rp=
3 0
+
45 0 0 + + -45 4.78 19.09
2025 0
ARD=
-5400 0
=
48 -21.13
0 477.3 -0.0317
ARD .D=
2025 0
-5400 0
0 477.3
-0.0127
=
4.182 0.681
0.0014
AR=
52.182 -20.449
AGUEDO TORRES ALEXANDER
UNASAM-FIC