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Analisis de deflexiones
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Ejercicio 8
Metodo de Carga Unitaria.
Calculo de Reacciones
Armadura Real
+ Fx = 0
35 − Rax = 0
Rax = 35 kN
+ Ma = 0
− 35 0.5 − 25 3 − 40 3 − 10 4.5 − 22 4.5 + Rex 6 = 0
Rey =713
12 kN
+ Fy = 0
Ray − 25 − 40 − 10 − 22 +713
12= 0
Ray =451
12
Armadura Virtual 1.
Armadura Virtual 2
+ Ma = 0
− 1 3 + Rey 6 = 0
Rey =1
2 kN
+ Fy = 0
Ray − 1 +1
2= 0
Ray =1
2kN
+ Ma = 0
− 1 1 + Rey 6 = 0
Rey =1
6 kN
Rax
+ Fy = 0
Ray +1
6= 0
Ray = −1
6kN
+ Fx = 0
1 − Rax = 0
Rax = 1 kN
1. Análisis de Nodos
Nodo A
AB
AH
451/12 kN
35 kN
AB
AH
1/2 kN
AB AH
1/6 kN
1 kN
Fx = 0
−35 − AB 3
10 + AH
6
37 = 0
AB =6 10
3 37 AH −
35 10
3
Fy = 0
451
12−
1
10 AB +
1
37 AH = 0
451
12−
6
3 37AH +
35
3+
1
37AH = 0
𝐀𝐇 = 𝟐𝟗𝟗.𝟓𝟖 𝐤𝐍
𝐀𝐁 = 𝟐𝟕𝟒.𝟓𝟗 𝐤𝐍
1
1
6
3
𝟏𝟎
Fx = 0
−AB 3
10 + AH
6
37 = 0
AH =3 37
6 10 AB
Fy = 0
1
2−
1
10 AB +
1
37 AH = 0
1
2−
1
10 AB +
3
6 10AB = 0
𝐀𝐁 = 𝟑.𝟏𝟔𝟐 𝐤𝐍
𝐀𝐇 = 𝟑.𝟎𝟑𝟗 𝐤𝐍
Fx = 0
−1− AB 3
10 + AH
6
37 = 0
AB =6 10
3 37 AH −
10
3
Fy = 0
−1
6−
1
10 AB +
1
37 AH = 0
−1
6−
6
3 37 AH +
1
3+
1
37AH = 0
𝐀𝐇 = 𝟏.𝟎𝟏𝟒 𝐤𝐍
𝐀𝐇 = 𝟎
AB
Nodo E
ED
EF
1/2 kN 1/6 kN
713/12 KN
ED EF
ED
EF
Fx = 0
ED 3
10 − EF
6
37 = 0
ED =6 10
3 37 EF
Fy = 0
713
12−
1
10 ED +
1
37 EF = 0
713
12−
6
3 37EF +
1
37EF = 0
𝐄𝐅 = 𝟑𝟔𝟏.𝟒𝟐 𝐤𝐍
𝐄𝐃 = 𝟑𝟕𝟓.𝟕𝟖 𝐤𝐍
Fx = 0
ED 3
10 − EF
6
37 = 0
ED =6 10
3 37 EF
Fy = 0
1
2−
1
10 ED +
1
37 EF = 0
1
2−
6
3 37EF +
1
37EF = 0
𝐄𝐅 = 𝟑.𝟎𝟑𝟗 𝐤𝐍
𝐄𝐃 = 𝟑.𝟏𝟔𝟐 𝐤𝐍
Fx = 0
ED 3
10 − EF
6
37 = 0
ED =6 10
3 37 EF
Fy = 0
1
6−
1
10 ED +
1
37 EF = 0
1
6−
6
3 37EF +
1
37EF = 0
𝐄𝐅 = 𝟏.𝟎𝟏𝟒 𝐤𝐍
𝐄𝐃 = 𝟏.𝟎𝟓𝟒 𝐤𝐍
Nodo F
FD
FG
FE
FG
FG
FD FD
FE FE
361.417kN 3.039kN
1.014kN
22 kN
Fx = 0
−FG 6
37 + FE
6
37 = 0
FG = FE
Fy = 0
−22 + FD − 1
37 FE +
1
37 FG = 0
−22 + FD = 0
𝐅𝐆 = 𝟑𝟔𝟏.𝟒𝟏𝟕 𝐤𝐍
𝐅𝐃 = 𝟐𝟐 𝐤𝐍
Fx = 0
−FG 6
37 + FE
6
37 = 0
FG = FE
Fy = 0
+FD − 1
37 FE +
1
37 FG = 0
FD = 0
𝐅𝐆 = 𝟑.𝟎𝟑𝟗 𝐤𝐍
𝐅𝐃 = 𝟎 𝐤𝐍
Fx = 0
−FG 6
37 + FE
6
37 = 0
FG = FE
Fy = 0
+FD − 1
37 FE +
1
37 FG = 0
FD = 0
𝐅𝐆 = 𝟏.𝟎𝟏𝟒 𝐤𝐍
𝐅𝐃 = 𝟎 𝐤𝐍
Nodo D
DG
DC DE
DC
DC
10
DE
DE
375.783kN 3.162kN
1.054kN
DG
22 kN
DG
Fy = 0
−10 − 22 − 1
10 DC +
1
10 DE = 0
𝐃𝐂 = 𝟐𝟕𝟒.𝟓𝟗 𝐤𝐍
Fx = 0
DG − DE 3
10 + DC
3
10 = 0
𝐃𝐆 = 𝟗𝟔 𝐤𝐍
Fy = 0
− 1
10 DC +
1
10 DE = 0
DC = DE
𝐃𝐂 = 𝟑.𝟏𝟔𝟐 𝐤𝐍
Fx = 0
DG − DE 3
10 + DC
3
10 = 0
𝐃𝐆 = 𝟎 𝐤𝐍
Fy = 0
− 1
10 DC +
1
10 DE = 0
DC = DE
𝐃𝐂 = 𝟏.𝟎𝟓𝟒 𝐤𝐍
Fx = 0
DG − DE 3
10 + DC
3
10 = 0
𝐃𝐆 = 𝟎 𝐤𝐍
Nodo H
HB
HA HG
HA HA
HB HB
HG HG
274.59kN 3.039kN
1.014kN
Fx = 0
−HA 6
37 + HG
6
37 = 0
HA = HG
Fy = 0
HB − 1
37 HA +
1
37 HG = 0
HB = 0
𝐇𝐆 = 𝟐𝟕𝟒.𝟓𝟗 𝐤𝐍
Fx = 0
−HA 6
37 + HG
6
37 = 0
HA = HG
Fy = 0
HB − 1
37 HA +
1
37 HG = 0
HB = 0
𝐇𝐆 = 𝟑.𝟎𝟑𝟗 𝐤𝐍
Fx = 0
−HA 6
37 + HG
6
37 = 0
HA = HG
Fy = 0
HB − 1
37 HA +
1
37 HG = 0
HB = 0
𝐇𝐆 = 𝟏.𝟎𝟏𝟒 𝐤𝐍
Nodo B
BC
BG
274.59 KN
Fy = 0
− 1
10 BC +
1
10 BA = 0
𝐁𝐂 = 𝟐𝟕𝟒.𝟓𝟗 𝐤𝐍
Fx = 0
35 − BG = 0
𝐁𝐆 = 𝟑𝟓 𝐤𝐍
35KN
Fy = 0
− 1
10 BC +
1
10 BA = 0
BC = BA
𝐁𝐂 = 𝟑.𝟏𝟔𝟐 𝐤𝐍
Fx = 0
𝐁𝐆 = 𝟎 𝐤𝐍
BA BA 3.162 KN
BC
BG
0 KN
BG
BA
BC
Fy = 0
− 1
10 BC +
1
10 BA = 0
BC = BA
𝐁𝐂 = 𝟎 𝐤𝐍
Fx = 0
𝐁𝐆 = 𝟎 𝐤𝐍
Nodo G
40 kN
299.58 KN
GC
96
361.417kN
Fy = 0
GC − 40 − 1
37 GH −
1
37 GF = 0
GC − 40 − 1
37 299.58 −
1
37 361.417 = 0
𝐆𝐂 = 𝟏𝟒𝟖.𝟔𝟔𝟕 𝐊𝐍
35
GD GB
GH GF
GC
3.039kN
GF
3.039 KN
GH
1.014 KN
GH
1.014kN
GF
GC
Fy = 0
GC − 1
37 GH −
1
37 GF = 0
GC − 1
37 3.039 2 = 0
𝐆𝐂 = 𝟏𝐊𝐍
Fy = 0
GC − 1
37 GH −
1
37 GF = 0
GC − 1
37 1.014 2 = 0
𝐆𝐂 = 𝟎.𝟑𝟑𝟑𝐊𝐍
TABLA DE RESULTADOS
Elemento A(cm3) L (cm) L/A P Vc Hc P.p .L/A P.p .L/A
AB 3 158.113 52.704 -274.59 -3.162 0 45760.4367 0
BC 3 158.113 52.704 -274.59 -3.162 0 45760.4367 0
CD 3 158.113 52.704 -274.59 -3.162 -1.054 45760.4367 15253.48
DE 3 158.113 52.704 -375.78 -3.162 -1.054 62623.755 20874.59
EF 2 152.069 76.034 361.42 3.039 1.014 83512.353 27864.93
FG 2 152.069 76.034 361.42 3.039 1.014 83512.353 27864.93
GH 2 152.069 76.034 299.58 3.039 1.014 69223.1495 23097.16
HA 2 152.069 76.034 299.58 3.039 1.014 69223.1495 23097.16
BH - 152.062 0 0 0 0 0
BG 3 150 50 -35 0 0 0 0
DG 3 150 50 -96 0 0 0 0
DF 2 152.062 76.034 22 0 0 0 0
CG 2 50 25 148.67 1 0.33 3716.75 1226.528 509092.82 139278.8
Dado un módulo de elasticidad de 200 GPa= 200x109 N/m2= 200,000 kN/cm2
1kN. =509092.82 kN
200 000 kN
= 2.545
1kN. H =139278.8 kN
200 000 kN
H = 0.696
