Download - Cap-II-AE-II.pdf
![Page 1: Cap-II-AE-II.pdf](https://reader033.vdocuments.co/reader033/viewer/2022051517/55cf8eb6550346703b94d1d2/html5/thumbnails/1.jpg)
LÍNEAS DE INFLUENCIA EN ESTRUCTURASISOSTÁTICAS... Continuación
! Líneas de Influencia para Trabes! Líneas de Influencia para Reacciones! Líneas de Influencia para Cortantes! Líneas de Influencia para Momentos Flectores! Valores Máximos! Líneas de Influencia para Armaduras! Líneas de Influencia para Reacciones! Líneas de Influencia para la Carga Axial! Valores Máximos
![Page 2: Cap-II-AE-II.pdf](https://reader033.vdocuments.co/reader033/viewer/2022051517/55cf8eb6550346703b94d1d2/html5/thumbnails/2.jpg)
A B C D EFGH
(b/L)P1 (a/L)P1 P2
P1
a b
P2
L
A B
P1
D
P2
A
B C D
FGH
Línea de Influencia para Trabes� Fuerzas aplicadas a la Trabe
![Page 3: Cap-II-AE-II.pdf](https://reader033.vdocuments.co/reader033/viewer/2022051517/55cf8eb6550346703b94d1d2/html5/thumbnails/3.jpg)
A FGH
1
0.5 0.5
RGRH
RH
1
� Reacción
![Page 4: Cap-II-AE-II.pdf](https://reader033.vdocuments.co/reader033/viewer/2022051517/55cf8eb6550346703b94d1d2/html5/thumbnails/4.jpg)
A
FGH
RG
1
0.5 0.5
RG
11
RH
![Page 5: Cap-II-AE-II.pdf](https://reader033.vdocuments.co/reader033/viewer/2022051517/55cf8eb6550346703b94d1d2/html5/thumbnails/5.jpg)
A
FGH
La b b b
1
0.5 0.5
1
1
VC
VC
VCx
2b/L
-(a+b)/L
1
C
� Cortante
![Page 6: Cap-II-AE-II.pdf](https://reader033.vdocuments.co/reader033/viewer/2022051517/55cf8eb6550346703b94d1d2/html5/thumbnails/6.jpg)
A
FGH
L
VCD
VCD
1
0.5 0.5
1
0.5 0.5
VCD x
-a/L
b/L
a b
![Page 7: Cap-II-AE-II.pdf](https://reader033.vdocuments.co/reader033/viewer/2022051517/55cf8eb6550346703b94d1d2/html5/thumbnails/7.jpg)
A
C
FGH
L
a b
ab/L
1
0.5 0.5
MCx
1
1
MCMC
C
� Momento Flector
![Page 8: Cap-II-AE-II.pdf](https://reader033.vdocuments.co/reader033/viewer/2022051517/55cf8eb6550346703b94d1d2/html5/thumbnails/8.jpg)
A
B C D E
FGH
L
MFx
1
0.5 0.5
1
0.5 0.5
MFMF
a b
ab/L
![Page 9: Cap-II-AE-II.pdf](https://reader033.vdocuments.co/reader033/viewer/2022051517/55cf8eb6550346703b94d1d2/html5/thumbnails/9.jpg)
1.5 m 1.5 m
A
C D
HF
3 m 3 m 3 m 3 m
G
Ejemplo 1
Dibujar las líneas de influencia para:- Las Reacciones RG y RF- Cortante VCD- Los Momentos Flectores MC y MH
![Page 10: Cap-II-AE-II.pdf](https://reader033.vdocuments.co/reader033/viewer/2022051517/55cf8eb6550346703b94d1d2/html5/thumbnails/10.jpg)
1
RG
x
RG
1.333
0.6670.333
1.5 m 1.5 m
A HF
3 m 3 m 3 m 3 m G
![Page 11: Cap-II-AE-II.pdf](https://reader033.vdocuments.co/reader033/viewer/2022051517/55cf8eb6550346703b94d1d2/html5/thumbnails/11.jpg)
RF
x
RF
10.667
0.333
-0.333
1.5 m 1.5 m
A HF
3 m 3 m 3 m 3 m G
![Page 12: Cap-II-AE-II.pdf](https://reader033.vdocuments.co/reader033/viewer/2022051517/55cf8eb6550346703b94d1d2/html5/thumbnails/12.jpg)
1.5 m 1.5 m
A HF
3 m 3 m 3 m 3 m G
VCD
x
VCD
VCD
1
0.5 0.5
4.5/9
-4.5/9
0.333
-0.333
0.333
C D
![Page 13: Cap-II-AE-II.pdf](https://reader033.vdocuments.co/reader033/viewer/2022051517/55cf8eb6550346703b94d1d2/html5/thumbnails/13.jpg)
1.5 m 1.5 m
A HF
3 m 3 m 3 m 3 m G
MCMC
1
1
MC
x
(3)(6)/9 = 2
-2
1
C
![Page 14: Cap-II-AE-II.pdf](https://reader033.vdocuments.co/reader033/viewer/2022051517/55cf8eb6550346703b94d1d2/html5/thumbnails/14.jpg)
MHMH
1
0.5 0.5
MH
x
(4.5)(4.5)/9 = 2.25
1.51.5
-1.5
1.5 m 1.5 m
A HF
3 m 3 m 3 m 3 m G
![Page 15: Cap-II-AE-II.pdf](https://reader033.vdocuments.co/reader033/viewer/2022051517/55cf8eb6550346703b94d1d2/html5/thumbnails/15.jpg)
2 m
AGB
6 @ 4 m = 24 m
C D E FH
Ejemplo 2
Dibujar las líneas de influencia para la trabe:- Las reacciones en C y G- Cortante en E y H- Momento Flector en H
![Page 16: Cap-II-AE-II.pdf](https://reader033.vdocuments.co/reader033/viewer/2022051517/55cf8eb6550346703b94d1d2/html5/thumbnails/16.jpg)
RC
RC
10.75 0.50 0.25
1.25
AGB
6 @ 4 m = 24 m
C D E FH
2 m
SOLUCIÓN
![Page 17: Cap-II-AE-II.pdf](https://reader033.vdocuments.co/reader033/viewer/2022051517/55cf8eb6550346703b94d1d2/html5/thumbnails/17.jpg)
AGB
6 @ 4 m = 24 m
C D E FH
2 m
RG
RG
1
1
10.25 0.50
0.75
-0.25
![Page 18: Cap-II-AE-II.pdf](https://reader033.vdocuments.co/reader033/viewer/2022051517/55cf8eb6550346703b94d1d2/html5/thumbnails/18.jpg)
AGB
6 @ 4 m = 24 m
C D E FH
2 m
VE
1
1
1
1
VE
VE
8/16 = 0.5
-8/16 = 0.5-0.25
0.250.25
8 m 8 m
![Page 19: Cap-II-AE-II.pdf](https://reader033.vdocuments.co/reader033/viewer/2022051517/55cf8eb6550346703b94d1d2/html5/thumbnails/19.jpg)
AGB
6 @ 4 m = 24 m
C D E FH
2 m
6/16 = 0.375
-10/16 = 0.625
VH
1
1
VH
VH
1
0.5 0.5
-0.25
0.250.25
-0.50
10 m 6 m
![Page 20: Cap-II-AE-II.pdf](https://reader033.vdocuments.co/reader033/viewer/2022051517/55cf8eb6550346703b94d1d2/html5/thumbnails/20.jpg)
AGB
6 @ 4 m = 24 m
C D E FH
2 m
MH
1
1
1
0.5 0.5
MHMH
(10)(6)/16 = 3.75
1.50 2.50
-1.50
3
10 m 6 m
![Page 21: Cap-II-AE-II.pdf](https://reader033.vdocuments.co/reader033/viewer/2022051517/55cf8eb6550346703b94d1d2/html5/thumbnails/21.jpg)
2 m 2 m 4 m 4 m 4 m
2 m 2 m
A
B C D
FGH
Ejemplo 3
Dibujar las líneas de influencia para:- Reacciones RG y RH- Cortantes VC y VCD- Momentos Flectores MC, MD y MF
Y determine los valores máximos para:- Reacciones (RG)max y (RH)max- Cortante (VCD)max
Debido a:- Una carga muerta uniforme de 2 kN/m- Una carga viva uniforme de 5 kN/m- Una carga viva concentrada de 50 kN
![Page 22: Cap-II-AE-II.pdf](https://reader033.vdocuments.co/reader033/viewer/2022051517/55cf8eb6550346703b94d1d2/html5/thumbnails/22.jpg)
2 m 2 m 4 m 4 m 4 m
2 m 2 m
A
B C D
FGH
RG
1
0.5 0.5
10/146/14
2/141/14
RG
11
RH
![Page 23: Cap-II-AE-II.pdf](https://reader033.vdocuments.co/reader033/viewer/2022051517/55cf8eb6550346703b94d1d2/html5/thumbnails/23.jpg)
2 m 2 m 4 m 4 m 4 m
2 m 2 m
A
B C D
E
FGH
1
0.5 0.5
RGRH
RH
1 12/148/14
4/146/14
![Page 24: Cap-II-AE-II.pdf](https://reader033.vdocuments.co/reader033/viewer/2022051517/55cf8eb6550346703b94d1d2/html5/thumbnails/24.jpg)
2 m 2 m 4 m 4 m 4 m
2 m 2 m
A
B C D
FGH
VC
VC
1
0.5 0.5
VCx
8/14
-6/14
1
1
1
-2/14
4/14
-1/14
![Page 25: Cap-II-AE-II.pdf](https://reader033.vdocuments.co/reader033/viewer/2022051517/55cf8eb6550346703b94d1d2/html5/thumbnails/25.jpg)
-8/14
6/14
2 m 2 m 4 m 4 m 4 m
2 m 2 m
A
B C D
FGH
VF
VF
1
0.5 0.5
1
0.5 0.5
VCD x-1/14 -2/14
-6/14
4/14
![Page 26: Cap-II-AE-II.pdf](https://reader033.vdocuments.co/reader033/viewer/2022051517/55cf8eb6550346703b94d1d2/html5/thumbnails/26.jpg)
2 m 2 m 4 m 4 m 4 m
2 m 2 m
A
B C D
FGH
(6)(8)/14 = 3.43
1
0.5 0.5
MCx
(4/8)(3.43)(2/6)(3.43)
1
1
MCMC
![Page 27: Cap-II-AE-II.pdf](https://reader033.vdocuments.co/reader033/viewer/2022051517/55cf8eb6550346703b94d1d2/html5/thumbnails/27.jpg)
2 m 2 m 4 m 4 m 4 m
2 m 2 m
A
B C D
E
FGH
1
0.5 0.5
1
1
MD x
(10)(4)/14 = 2.86(6/10)(2.86)(2/10)(2.86)
MDMD
![Page 28: Cap-II-AE-II.pdf](https://reader033.vdocuments.co/reader033/viewer/2022051517/55cf8eb6550346703b94d1d2/html5/thumbnails/28.jpg)
(8)(6)/14=3.428
2 m 2 m 4 m 4 m 4 m
2 m 2 m
A
B C D
FGH
MFx
1
0.5 0.5
1
0.5 0.5
2.2852.5710.8570.429
MFMF
![Page 29: Cap-II-AE-II.pdf](https://reader033.vdocuments.co/reader033/viewer/2022051517/55cf8eb6550346703b94d1d2/html5/thumbnails/29.jpg)
[0.5×4 (2/14)] + [(0.5)(2/14 + 1)(12) = 7.143RG
110/14
6/142/14
1/14
2 kN/m5 kN/m
50 kN
(RG)max = (2)(7.143) + (5)(7.143) + (50)(1)
= 100 kN
2 m 2 m 4 m 4 m 4 m
2 m 2 m
A
B C D E
FGH
Reacción Máxima
![Page 30: Cap-II-AE-II.pdf](https://reader033.vdocuments.co/reader033/viewer/2022051517/55cf8eb6550346703b94d1d2/html5/thumbnails/30.jpg)
2 m 2 m 4 m 4 m 4 m
2 m 2 m
A
B C D E
FGH
(0.5)(16)(12/14) = 6.857RH
12/148/14
4/146/14
2 kN/m5 kN/m
50 kN
(RH)max = (2)(6.857) + (5)(6.857) + (50)(12/14)
= 90.86 kN
Reacción Máxima
![Page 31: Cap-II-AE-II.pdf](https://reader033.vdocuments.co/reader033/viewer/2022051517/55cf8eb6550346703b94d1d2/html5/thumbnails/31.jpg)
2 m 2 m 4 m 4 m 4 m
2 m 2 m
A
B C D E
FGH
+ (0.5)(5.6)(4/14) = + 0.8
-[0.5(4)(2/14) + 0.5(2/14) + (6/14)(4) + 0.5(2.4)(6/14)] = -1.943
2.4 mVCD
-1/14 -2/14-6/14
4/14
2 kN/m
5 kN/m50 kN
(VCD)max = (2)(-1.943 + 0.8) (5)(-1.943) + (50)(-6/14)
= -33.43 kN
C D
Cortante Máximo
![Page 32: Cap-II-AE-II.pdf](https://reader033.vdocuments.co/reader033/viewer/2022051517/55cf8eb6550346703b94d1d2/html5/thumbnails/32.jpg)
2 m 2 m 2 m 2 m 2 m
A B E C D
Tarea
Dibujar las líneas de influencia par:- Las reacciones de los apoyos A y D- La cortante y el momento flector en E- El momento flector en B y C .
![Page 33: Cap-II-AE-II.pdf](https://reader033.vdocuments.co/reader033/viewer/2022051517/55cf8eb6550346703b94d1d2/html5/thumbnails/33.jpg)
RA
x
1
RE
x
1
(3/4)(2/4)
(1/4)
(2/4)(3/4)
(1/4)
AB C
E
GH
D
F
4 @ x m
h
Línea de Influencia para Armaduras
![Page 34: Cap-II-AE-II.pdf](https://reader033.vdocuments.co/reader033/viewer/2022051517/55cf8eb6550346703b94d1d2/html5/thumbnails/34.jpg)
AB C
E
GH
D
F
4 @ x m
h
1 kN
FCB
RE
FGB
FGH
FBC x
(2x/h)(1/4)
+ ΣMG = 0:
0)()2( =− hFxR BCE
EBC RhxF 2
=
RE
x
1 (2/4)
(3/4)
(1/4)
Considerar la sección a la derecha
![Page 35: Cap-II-AE-II.pdf](https://reader033.vdocuments.co/reader033/viewer/2022051517/55cf8eb6550346703b94d1d2/html5/thumbnails/35.jpg)
AB C
E
GH
D
F
4 @ x m
h
1 kN
FBC
RA
FBG
+ ΣMG = 0:
0)2()( =− xRhF ABC
ABC RhxF 2
=
FBC x
(2x/h)(1/4)
RA
x
1 (3/4)
(2/4)(1/4)
(2x/h)(1/4)
(2x/h)(2/4)
Considere la sección a la izquierdaFHG
![Page 36: Cap-II-AE-II.pdf](https://reader033.vdocuments.co/reader033/viewer/2022051517/55cf8eb6550346703b94d1d2/html5/thumbnails/36.jpg)
4 kN 1.5 kN2 m
AB C
E
GH
4 m
D
F
3 m 3 m 3 m 3 m
Ejemplo 4
Determinar la fuerza máxima desarrollada enlos miembros BC, BG y CG de la armaduradebido a las cargas de las ruedas del vehículo.Suponga que las cargas se aplican directamentea la armadura.Vehículo
![Page 37: Cap-II-AE-II.pdf](https://reader033.vdocuments.co/reader033/viewer/2022051517/55cf8eb6550346703b94d1d2/html5/thumbnails/37.jpg)
RA
x
1 0.75
0.50.25
RE
x
0.50.75
0.25
1
AB C
E
GH
4 m
D
F
3 m 3 m 3 m 3 m
SOLUCIÓN
![Page 38: Cap-II-AE-II.pdf](https://reader033.vdocuments.co/reader033/viewer/2022051517/55cf8eb6550346703b94d1d2/html5/thumbnails/38.jpg)
AB C
E
GH
4 m
D
F
3 m 3 m 3 m 3 m1 kN
FBC = (6/4) RE = 1.5 RE
FCB
RE
FGB
FGH
RE
x
0.50.75
0.25
1
FBC x
1.5(0.25) = 0.375
+ ΣMG = 0:
Considerar la sección de la derechaLínea de Inluencia de FBC
![Page 39: Cap-II-AE-II.pdf](https://reader033.vdocuments.co/reader033/viewer/2022051517/55cf8eb6550346703b94d1d2/html5/thumbnails/39.jpg)
AB C
E
GH
4 m
D
F
3 m 3 m 3 m 3 m
0.375
1.5(0.5) = 0.75
1 kN
FBC
RA
FBG
RA
x
1 0.75
0.50.25
FBC = 1.5 RA
FBC
x
0.375
+ ΣMG = 0:
Considerar la sección de la izquierdaFHG
![Page 40: Cap-II-AE-II.pdf](https://reader033.vdocuments.co/reader033/viewer/2022051517/55cf8eb6550346703b94d1d2/html5/thumbnails/40.jpg)
AB C
E
GH
4 m
D
F
3 m 3 m 3 m 3 m
RE
x
0.50.75
0.25
1
1 kN
ΣFy = 0 ;
FBG cos θ = RE
FGH
FGB
FCB
RE
θ
FBG = 1.25 RE
FBG (4/5) = RE1.25(0.25) = 0.3125
FBG x
Línea de Inffluencia para FBGConsiderar la sección de la derecha
![Page 41: Cap-II-AE-II.pdf](https://reader033.vdocuments.co/reader033/viewer/2022051517/55cf8eb6550346703b94d1d2/html5/thumbnails/41.jpg)
AB C
E
GH
4 m
D
F
3 m 3 m 3 m 3 m1 kN
ΣFy = 0 ;
FBG cos θ + RA = 0RA
x
1 0.75
0.50.25
FHG
FBC
RA
FBGθ
0.3125
FBG x
FBG = -1.25 RA
FBG (4/5) = -RA
-0.3125-1.25(0.5) = -0.625
Considerar la sección de la izquierda
![Page 42: Cap-II-AE-II.pdf](https://reader033.vdocuments.co/reader033/viewer/2022051517/55cf8eb6550346703b94d1d2/html5/thumbnails/42.jpg)
AB C
E
GH
4 m
D
F
3 m 3 m 3 m 3 m
FCG x
1 kN
0
1 kN
0
Línea de Influencia para FCG
![Page 43: Cap-II-AE-II.pdf](https://reader033.vdocuments.co/reader033/viewer/2022051517/55cf8eb6550346703b94d1d2/html5/thumbnails/43.jpg)
AB C
E
GH
4 m
D
F
3 m 3 m 3 m 3 m
FCG x
1 kN
1
1
![Page 44: Cap-II-AE-II.pdf](https://reader033.vdocuments.co/reader033/viewer/2022051517/55cf8eb6550346703b94d1d2/html5/thumbnails/44.jpg)
AB C
E
GH
4 m
D
F
3 m 3 m 3 m 3 m
(FBC)max = (4)(0.75) + (1.5)(0.5)
= 3.75 kN (T)
0.375
0.75FBC
x
0.375
4 kN 1.5 kN2 m
0.5
(FBC)máx por las cargas del vehículo
![Page 45: Cap-II-AE-II.pdf](https://reader033.vdocuments.co/reader033/viewer/2022051517/55cf8eb6550346703b94d1d2/html5/thumbnails/45.jpg)
AB C
E
GH
4 m
D
F
3 m 3 m 3 m 3 m
(FBG)max = (4)(-0.625) + (1.5)(-0.417) = -3.126 kN (C)
FBG x
-0.3125
0.3125
-0.625
4 kN 1.5 kN2 m
-0.417
(FBG)máx por cargas del vehículo
![Page 46: Cap-II-AE-II.pdf](https://reader033.vdocuments.co/reader033/viewer/2022051517/55cf8eb6550346703b94d1d2/html5/thumbnails/46.jpg)
AB C
E
GH
4 m
D
F
3 m 3 m 3 m 3 m
FCG x
1
0.333
(FCG)max = (4)(1) + (1.5)(0.333) = 4.50 kN (T)
4 kN 1.5 kN2 m
(FCG)máx por cargas del vehículo
![Page 47: Cap-II-AE-II.pdf](https://reader033.vdocuments.co/reader033/viewer/2022051517/55cf8eb6550346703b94d1d2/html5/thumbnails/47.jpg)
AB C
E
HI
4 m
D
F
3 m 3 m 3 m 3 m
GJ
Ejemplo 5
Dibujar las líneas de influencia de FBC , FBH y FCH.
![Page 48: Cap-II-AE-II.pdf](https://reader033.vdocuments.co/reader033/viewer/2022051517/55cf8eb6550346703b94d1d2/html5/thumbnails/48.jpg)
RA
x
1 0.75
0.50.25
RE
x
0.50.75
0.25
1
AB C
E
HI
4 m
D
F
3 m 3 m 3 m 3 m
GJSOLUCIÓN
![Page 49: Cap-II-AE-II.pdf](https://reader033.vdocuments.co/reader033/viewer/2022051517/55cf8eb6550346703b94d1d2/html5/thumbnails/49.jpg)
AB C
E
HI
4 m
D
G
3 m 3 m 3 m 3 m
FJ
1 kN
FBC = (6/4) RE = 1.5 RE
FCB
RE
FHB
FHI
RE
x
0.50.75
0.25
1
FBC x
1.5(0.25) = 0.375
+ ΣMH = 0:
Línea de Influencia de FBCConsiderar la sección de la derecha
![Page 50: Cap-II-AE-II.pdf](https://reader033.vdocuments.co/reader033/viewer/2022051517/55cf8eb6550346703b94d1d2/html5/thumbnails/50.jpg)
AB C
E
HI
4 m
D
G
3 m 3 m 3 m 3 m
FJ
0.375
1.5(0.5) = 0.75
1 kN
FBC
RA
FBH
RA
x
1 0.75
0.50.25
FBC = 1.5 RA
FBC
x
0.375
+ ΣMH = 0:
Considerar la sección de la izquierdaFHI
![Page 51: Cap-II-AE-II.pdf](https://reader033.vdocuments.co/reader033/viewer/2022051517/55cf8eb6550346703b94d1d2/html5/thumbnails/51.jpg)
AB C
E
HI
4 m
D
G
3 m 3 m 3 m 3 m
FJ
RE
x
0.50.75
0.25
1
1 kN
ΣFy = 0 ;
FBH cos θ = RE
FHI
FBH
FCB
RE
θ
FBH = 1.25 RE
FBH (4/5) = RE1.25(0.25) = 0.3125
FBH x
Línea de Influencia de FBHConsiderar la sección de la derecha
![Page 52: Cap-II-AE-II.pdf](https://reader033.vdocuments.co/reader033/viewer/2022051517/55cf8eb6550346703b94d1d2/html5/thumbnails/52.jpg)
AB C
E
HI
4 m
D
G
3 m 3 m 3 m 3 m
FJ
1 kN
ΣFy = 0 ;
FBH cos θ + RA = 0RA
x
1 0.75
0.50.25
FHI
FBC
RA
FBHθ
0.3125
FBH x
FBH = -1.25 RA
FBH (4/5) = -RA
-0.3125-1.25(0.5) = -0.625
Considerar la sección de la izquierda
![Page 53: Cap-II-AE-II.pdf](https://reader033.vdocuments.co/reader033/viewer/2022051517/55cf8eb6550346703b94d1d2/html5/thumbnails/53.jpg)
AB C
E
HI
4 m
D
G
3 m 3 m 3 m 3 m
FJ
FCH x
1 kN
FGH
FCH
FCB
RE
x
0.50.75
0.25
1
REΣFy = 0 ;FCH = -RE
-0.25
Línea de Influencia de FCH
![Page 54: Cap-II-AE-II.pdf](https://reader033.vdocuments.co/reader033/viewer/2022051517/55cf8eb6550346703b94d1d2/html5/thumbnails/54.jpg)
AB C
E
HI
4 m
D
G
3 m 3 m 3 m 3 m
FJ
FCH x
1 kN
FGH
FCH
FCBRA
ΣFy = 0 ;FCH = RA
-0.25
RA
x
1 0.75
0.50.25
0.50.25
![Page 55: Cap-II-AE-II.pdf](https://reader033.vdocuments.co/reader033/viewer/2022051517/55cf8eb6550346703b94d1d2/html5/thumbnails/55.jpg)
AB C
E
G
H
3 m
D
F
3 m 3 m 3 m 3 m
1.5 m
Ejemplo 6
Dibujar las líneas de influencia para FBC , FBG y FHG.
![Page 56: Cap-II-AE-II.pdf](https://reader033.vdocuments.co/reader033/viewer/2022051517/55cf8eb6550346703b94d1d2/html5/thumbnails/56.jpg)
RA
x
1 0.75
0.50.25
RE
x
0.50.75
0.25
1
AB C
E
G
H
3 m
D
F
3 m 3 m 3 m 3 m
1.5 m
SOLUCIÓN
![Page 57: Cap-II-AE-II.pdf](https://reader033.vdocuments.co/reader033/viewer/2022051517/55cf8eb6550346703b94d1d2/html5/thumbnails/57.jpg)
AB C
E
G
H
3 m
D
F
3 m 3 m 3 m 3 m
1.5 m
1 kN
FBC = (6/4.5) RE = 1.33 RE
RE
x
0.50.75
0.25
1
FBC x
1.33(0.25) = 0.333
+ ΣMG = 0:
RE
FBC
FBG
FHGLínea de Influencia de FBC
Considerar la sección de la derecha
![Page 58: Cap-II-AE-II.pdf](https://reader033.vdocuments.co/reader033/viewer/2022051517/55cf8eb6550346703b94d1d2/html5/thumbnails/58.jpg)
AB C
E
G
H
3 m
D
F
3 m 3 m 3 m 3 m
1.5 m
1 kN
FBC = (6/4.5) RA = 1.33 RA
FBC x
1.33(0.25) = 0.333
+ ΣMG = 0:
RA
RA
x
1 0.75
0.50.25
FBC
FBG
FHG
1.33(0.5) = 0.667
1.33(0.25) = 0.333
Considerar la sección de la izquierda
![Page 59: Cap-II-AE-II.pdf](https://reader033.vdocuments.co/reader033/viewer/2022051517/55cf8eb6550346703b94d1d2/html5/thumbnails/59.jpg)
AB C
E
G
H
3 m
D
F
3 m 3 m 3 m 3 m
1.5 m
1 kN
RE
x
0.50.75
0.25
1
FHG x
+ ΣMB = 0:
RE
FBC
FBG
FHG
26.57o
(FHG cos 26.57)(3) = -9REFHG = -3.35RE
-3.35(0.25) = -0.84
Considerar la sección de la derechaLínea de Influencia de FHG
![Page 60: Cap-II-AE-II.pdf](https://reader033.vdocuments.co/reader033/viewer/2022051517/55cf8eb6550346703b94d1d2/html5/thumbnails/60.jpg)
AB C
E
G
H
3 m
D
F
3 m 3 m 3 m 3 m
1.5 m
1 kN
RA
RA
x
1 0.75
0.50.25
FBC
FBG
FHG
FHG x
-0.84
+ ΣMB = 0:
(FHG cos 26.57)(3) = -3RAFHG = -1.12RA
-1.12(0.5) = -0.56-1.12(0.25) = -0.28
Considerar la sección de la izquierda
![Page 61: Cap-II-AE-II.pdf](https://reader033.vdocuments.co/reader033/viewer/2022051517/55cf8eb6550346703b94d1d2/html5/thumbnails/61.jpg)
AB C
E
G
H
3 m
D
F
3 m 3 m 3 m 3 m
1.5 m
1 kN
RE
x
0.50.75
0.25
1
FBG x
RE
FBC
FBG
FHG
26.57o
FHG x
-0.84-0.56
-0.28
33.69o
1.2(0.25)-0.54(-0.84) = 0.75
Línea de Influencia de FBGConsiderar la sección de la derecha
+ ΣFy = 0:
-FBG cos 33.69- FHGsen26.57 + RE = 0FHG = +1.2RE - 0.54 FHG
![Page 62: Cap-II-AE-II.pdf](https://reader033.vdocuments.co/reader033/viewer/2022051517/55cf8eb6550346703b94d1d2/html5/thumbnails/62.jpg)
FHG x
-0.84-0.56
-0.28
AB C
E
G
H
3 m
D
F
3 m 3 m 3 m 3 m
1.5 m
1 kN
RA
FBC
FBG
FHG
RA
x
1 0.75
0.50.25
FBG x
0.75
-1.2(0. 5)-0.54(-0.56) = -0.30 -1.2(0. 25)-0.54(-0.28) = -0.15
Considerar la sección de la izquierda
+ ΣFy = 0:
RA + FBG cos33.69+ FHG sen 26.57 = 0
FBG = -1.2RA - 0.54FHG