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Problema 1:
Grafica:
Solución:
Viga liberada:
Calculo de reacciones:
∑M 5=0
−18∗19.8∗13.2+ 12∗9∗19.8∗13.2+6.6 X 4−13.2 X 3−19.8 X2+26.4 R1−M 1+M 5=0
R1=267320
−34X 2− X 3
2− X 44
+ M 126.4
− M 526.4
∑M 1=0
18∗19.8∗13.2−12∗9∗19.8∗13.2−6.6 X 2−13.2 X 3−19.8 X−26.4 R5−M 1+M 5=0
R5=267320
− X 24
− X32
−34X 4− M 1
26.4+ M 526.4
Calculo de momentos:
M 16=(267320 −34X 2− X 3
2− X 44
+ M 126.4
− M 526.4 )X−M 1
M 62=( 267320 −34X 2− X3
2− X 44
+ M 126.4
− M 526.4 )X−M 1−9 (x−3.3 )2+ 5
33( x−3.3 )3
M 23=(267320 − 34X2− X3
2−X 44
+ M 126.4
− M 526.4 )X−M 1−9 ( x−3.3 )2+ 5
33( x−3.3 )3+X 2(x−6.6)
M 57=(267320 − X24
− X 32
−34X 4− M 1
26.4+ M 526.4 )X−M 5
M 74=( 267320 − X 24
− X 32
−34X 4− M 1
26.4+ M 526.4 )X−M 5−9 ( x−3.3 )2+ 5
33( x−3.3 )3
M 43=( 267320 − X 24
−X 32
−34X 4− M 1
26.4+ M 526.4 )X−M 5−9 ( x−3.3 )2+ 5
33(x−3.3 )3+X 4 (x−6.6)
∂M /∂M 1 ∂M /∂ X 2 ∂M /∂ X 3 ∂M /∂ X 4 ∂M /∂M 5
M16 X26.4
−1 −34X
−X2
−X4
−X26.4
M62 X26.4
−1 −34X
−X2
−X4
−X26.4
M23 X26.4
−1 X4
−6.6 −X2
−X4
−X26.4
M57 −X26.4
−X4
−X2
−34X
X26.4
−1
M74 −X26.4
−X4
−X2
−34X
X26.4
−1
M43 −X26.4
−X4
−X2
X4
−6.6 X26.4
−1
Ɵ 1=∫MEI
∗∂M
∂M 1dx=0
Ɵ1= 14.5 EI
∫0
3.3
(( 267320 −34X 2− X 3
2− X 44
+ M 126.4
− M 526.4 )X−M 1)( X26.4−1)dx+¿
14.5EI
∫3.3
6.6
(( 267320 −34X 2− X 3
2− X 44
+ M 126.4
− M 526.4 )X−M 1−9 ( x−3.3 )2+ 5
33(x−3.3 )3)( X
26.4−1)dx+¿
13.25EI
∫6.6
13.2
(( 267320 −34X 2− X 3
2− X 44
+ M 126.4
− M 526.4 )X−M 1−9 ( x−3.3 )2+ 5
33(x−3.3 )3+X 2(x−6.6))( X26.4−1)dx+¿
14.5EI
∫0
3.3
(( 267320 − X 24
−X 32
−34X 4− M 1
26.4+ M 526.4 )X−M 5)(−X26.4 )dx+¿
14.5EI
∫3.3
6.6
(( 267320 − X 24
− X32
−34X 4− M 1
26.4+ M 526.4 )X−M 5−9 ( x−3.3 )2+ 5
33( x−3.3 )3)(−X26.4 )dx+¿
13.25EI
∫6.6
13.2
(( 267320 − X 24
− X32
− 34X 4− M 1
26.4+ M 526.4 )X−M 5−9 (x−3.3 )2+ 5
33( x−3.3 )3+X 4 (x−6.6))( −X
26.4 )dx=0−239520105+211640M 1+981552 X 2+1167408X 3+726000 X 4+115720M 5=0
∆2=∫MEI
∗∂M
∂ X 2dx=0
∆2= 14.5 EI
∫0
3.3
(( 267320 −34X 2− X 3
2− X 44
+ M 126.4
− M 526.4 )X−M 1)(−34 X)dx+¿
14.5 EI
∫3.3
6.6
(( 267320 −34X 2− X 3
2− X 44
+ M 126.4
− M 526.4 )X−M 1−9 ( x−3.3 )2+ 5
33(x−3.3 )3)(−34 X )dx+¿
13.25EI
∫6.6
13.2
(( 267320 −34X 2− X 3
2− X 44
+ M 126.4
− M 526.4 )X−M 1−9 ( x−3.3 )2+ 5
33(x−3.3 )3+X 2(x−6.6))( X4 −6.6)dx+¿
14.5 EI
∫0
3.3
(( 267320 − X 24
−X 32
−34X 4− M 1
26.4+ M 526.4 )X−M 5)(−X4 )dx+¿
14.5 EI
∫3.3
6.6
(( 267320 − X 24
− X32
−34X 4− M 1
26.4+ M 526.4 )X−M 5−9 ( x−3.3 )2+ 5
33( x−3.3 )3)(−X4 )dx+¿
13.25EI
∫6.6
13.2
(( 267320 − X 24
− X32
− 34X 4− M 1
26.4+ M 526.4 )X−M 5−9 (x−3.3 )2+ 5
33( x−3.3 )3+X 4 (x−6.6))(−X4 )dx=0
−120784867929+81796000M 1+477562800 X 2+600547200 X 3+378536400 X 4+60500000M 5=0
∆3=∫MEI
∗∂M
∂ X 3dx=0
∆3= 14.5EI
∫0
3.3
(( 267320 −34X 2− X 3
2− X 44
+ M 126.4
− M 526.4 )X−M 1)(−X2 )dx+¿
14.5EI
∫3.3
6.6
(( 267320 −34X 2− X 3
2− X 44
+ M 126.4
− M 526.4 )X−M 1−9 ( x−3.3 )2+ 5
33(x−3.3 )3)(−X2 )dx+¿
13.25EI
∫6.6
13.2
(( 267320 −34X 2− X 3
2− X 44
+ M 126.4
− M 526.4 )X−M 1−9 ( x−3.3 )2+ 5
33(x−3.3 )3+X 2(x−6.6))(−X2 )dx+¿
14.5EI
∫0
3.3
(( 267320 − X 24
−X 32
−34X 4− M 1
26.4+ M 526.4 )X−M 5)(−X2 )dx+¿
14.5EI
∫3.3
6.6
(( 267320 − X 24
− X32
−34X 4− M 1
26.4+ M 526.4 )X−M 5−9 ( x−3.3 )2+ 5
33( x−3.3 )3)(−X2 )dx+¿
13.25EI
∫6.6
13.2
(( 267320 − X 24
− X32
− 34X 4− M 1
26.4+ M 526.4 )X−M 5−9 (x−3.3 )2+ 5
33( x−3.3 )3+X 4 (x−6.6))(−X2 )dx=0
−57547607139+32428000M 1+200182400 X2+296014400 X3+200182400 X 4+32428000M 5=0
∆ 4=∫MEI
∗∂M
∂ X 4dx=0
∆ 4= 14.5 EI
∫0
3.3
(( 267320 −34X 2− X 3
2− X 44
+ M 126.4
− M 526.4 )X−M 1)(−X4 )dx+¿
14.5EI
∫3.3
6.6
(( 267320 −34X 2− X 3
2− X 44
+ M 126.4
− M 526.4 )X−M 1−9 ( x−3.3 )2+ 5
33(x−3.3 )3)(−X4 )dx+¿
13.25EI
∫6.6
13.2
(( 267320 −34X 2− X 3
2− X 44
+ M 126.4
− M 526.4 )X−M 1−9 ( x−3.3 )2+ 5
33(x−3.3 )3+X 2(x−6.6))(−X4 )dx+¿
14.5EI
∫0
3.3
(( 267320 − X 24
−X 32
−34X 4− M 1
26.4+ M 526.4 )X−M 5)(−34 X )dx+¿
14.5EI
∫3.3
6.6
(( 267320 − X 24
− X32
−34X 4− M 1
26.4+ M 526.4 )X−M 5−9 ( x−3.3 )2+ 5
33( x−3.3 )3)(−34 X )dx+¿
13.25EI
∫6.6
13.2
(( 267320 − X 24
− X32
− 34X 4− M 1
26.4+ M 526.4 )X−M 5−9 (x−3.3 )2+ 5
33( x−3.3 )3+X 4 (x−6.6))( X4 −6.6)dx=0
−120784867929+60500000M 1+378536400 X2+600547200 X3+477562800 X 4+81796000M 5=0
Ɵ 5=∫MEI
∗∂M
∂M 5dx=0
Ɵ5= 14.5 EI
∫0
3.3
(( 267320 −34X 2− X 3
2− X 44
+ M 126.4
− M 526.4 )X−M 1)( −X
26.4 )dx+¿ 1
4.5EI∫3.3
6.6
(( 267320 −34X 2− X 3
2− X 44
+ M 126.4
− M 526.4 )X−M 1−9 ( x−3.3 )2+ 5
33(x−3.3 )3)( −X
26.4 )dx+¿
13.25EI
∫6.6
13.2
(( 267320 −34X 2− X 3
2− X 44
+ M 126.4
− M 526.4 )X−M 1−9 ( x−3.3 )2+ 5
33(x−3.3 )3+X 2(x−6.6))( −X
26.4 )dx+¿
14.5EI
∫0
3.3
(( 267320 − X 24
−X 32
−34X 4− M 1
26.4+ M 526.4 )X−M 5)( X26.4−1)dx+¿
14.5EI
∫3.3
6.6
(( 267320 − X 24
− X32
−34X 4− M 1
26.4+ M 526.4 )X−M 5−9 ( x−3.3 )2+ 5
33( x−3.3 )3)( X26.4−1)dx+¿
13.25EI
∫6.6
13.2
(( 267320 − X 24
− X32
− 34X 4− M 1
26.4+ M 526.4 )X−M 5−9 (x−3.3 )2+ 5
33( x−3.3 )3+X 4 (x−6.6))( X26.4−1)dx=0
−239520105+115720M 1+726000 X 2+1167408X 3+981552X 4+211640M 5=0
SISTEMA DE ECUACIONES:
211640M 1+981552 X 2+1167408X 3+726000 X 4+115720M 5=239520105
81796000M 1+477562800 X 2+600547200 X 3+378536400 X 4+60500000M 5=120784867929
32428000M 1+200182400 X 2+296014400 X 3+200182400 X 4+32428000M 5=57547607139
60500000M 1+378536400 X2+600547200 X3+477562800 X 4+81796000M 5=120784867929
115720M 1+726000 X 2+1167408X 3+981552 X 4+211640M 5=239520105
M 1=113582762000
=18.3197903226
X 2=868121199200
=87.5122076613
X 3=357281149600
=72.0324798387
X 4=868121199200
=87.5122076613
M 5=113582762000
=18.3197903226
LAS ECUACIONES HIPERESTÁTICAS:
M 1=18.3200T .m⤹
X 2=87.5122T ↑
X 3=72.0325T↑
X 4=87.5122T ↑
M 1=18.3200T .m⤸
LAS OTRAS REACCIONES:
∑M 5=0
18.3200+87.5122∗6.6+72.0325∗13.2+87.5122∗19.8−18.3200−267.3∗13.2+R1∗26.4=0
R1=50202949600
=10.1215524194
∑M 1=0
−183200−87.5122∗6.6−72.0325∗13.2−87.5122∗19.8+18.3200+267.3∗13.2+R5∗26.4=0
R5=50202949600
=10.1215524194
DIAGRAMAS DE V Y M:
EJERCICIO 2:
GRAFICA
SOLUCION:
1) ACCIONES HIPERSTÁTICAS:
CALCULO DE REACCIONES:
∑M 5=0
R1 y∗21−X1∗10.5−100∗5.25=0
R14=25+ X 12
∑M 1=0
−R5 y∗21+X1∗10.5+100∗15.75=0
R5 y=75+ X 12
∑ FX=0
R1x=0
FUERZAS EN LAS BARRAS:
NUDO1:
∑ FY=0∑ FX=0
25+ X 12
+ 45F12=0 3
5 (−1254 −58X1)+F18=0
F12=−1254
−58X 1F18=75
4+ 38X1
NUDO 8:
∑ FY=0∑ FX=0
F82+ 45X 2=0−( 754 + 3
8X 1)+ 35 X 2+F 87=0
F82=−45X 2F 87=75
4+38X 1−3
5X2
NUDO 2:
∑ FY=0∑ FX=0
45 ( 1254 +5
8X 1)+ 45 X 2−45 F27=0 35 ( 1254 +5
8X 1)+ 35 ( 1254 + 5
8X 1+X 2)+F 23=0
F27=1254
+ 58X 1+X 2F23=−75
2−34X 1−3
5X 2
NUDO 3:
F82X2
F87F18
F27F82
F23
F12
F23 F34
X2
∑ FY=0∑ FX=0
−45X 1−4
5X 3−F37=0 F34−F23−3
5X 2+3
5X 3=0
F37=−45X 1−4
5X 3 F34=−75
2−34X 1−3
5X 3
X2
NUDO 7:
∑ FY=0∑ FX=0
F37−X 1+ 45F27+ 4
5F74=0F76−F 87−3
5F 27+3
5F74=0
F74=−1254
−58X 1+X 3 F76=225
4+ 38X1−3
5X 3
NUDO 6:
∑ FY=0∑ FX=0
F64−100+ 45X3=0 F65−F76−3
5X 3=0
F64=100−45X 3F 65=225
4+38X 1
NUDO 4:
∑ FX=0
X1
F27F37
F74
F76F87
100
X3F64
F65F76
F45F64
F34
F74
45 ( 1254 −5
8X1−X 3)−45 (100−45 X 3)−45 F 45=0F 45=−375
4−58X1
513.326787339+9.0562386157 X1+2.65425774135 X2+2.65425774125 X 3=0… (1 )
132.712887067+2.654257741 X 1+12.10652799 X 2+1.8064516129 X 3=0… (2 )
−358.51933868+2.65425774135 X 1+1.8064516129 X 2+12.1065279981 X 3=0… (3 )
RESOLVIENDO (1), (2) Y (3):
X 1=−69.1744818002R1=69.17448T ↑
X 2=−2.53428430977F 38=−2.53428T
X 3=−45.1578111054 F36=45.15781T
B) CALCULO DE LAS FUERZAS EN LAS BARRAS:
F12=11.98405T
F18=−7.19043T
F23=15.90143T
F27=−14.51834T
F28=2.02743T
F34=−12.71383T
F37=−34.09882T
F 45=−50.51595T
F 46=83.87775T
F 47=−29.32624T
F56=30.30957T
F67=3.214884T
F78=−5.66986T
F36=45.15781T
F38=−2.53428T