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