1

ESCUELA DE INGENIERIA CIVIL

ANALISIS ESTRUCTURAL

TEMA:ACTIVIDAD EN CLASES

Vigas y cargas puntuales “Método de Ecuaciones”

NOMBRE:ROSA GRANDA CASTILLO

AÑO:TERCERO “B”

DOCENTE:Ing. ALEXIS MARTÍNEZ

2011

2

Resolver la siguiente Viga.

Diagrama de Cortante

Diagrama de Momentos

∑ Fx=0

Ax=0

∑M=0

By . (6 )−152 ( 1

3.6)−5. ( 4 )=0

By=356

4 m 2 m

5 Tn52 Tn/m

∑ Fy=0

Ay+By=252

Donde:

Ay=203

152 Tn

203 Tn 35

6 Tn

52 h

x6

Si:526

=hx

∴ h=512

x

F=x .

512

x

2=

5 x2

24

5

356

203 Tn

1009

TRAMO I 0 ≤ x < 2

V=−356

−5 x2

24

M=35 x6

−5 x3

72Cuando:

x=0V=356;M=0

x=2V=5 ;M=1009

TRAMO II 2 ≤ x < 6

V=−356

−5−5 x2

24

M=35 x6

−5 ( x−2 )−5 x3

72Cuando:

x=2V=0 ;M=1009

x=6V=−203

;M=¿0

3

Resolver la siguiente Carga Puntual.

Donde:

W i=1

2 EI∫M2dx

W i=1

2 EI [∫0L8

(3Px )2dx+∫L8

L4

(2 Px+ PL8 )

2

dx+∫L4

L2

(Px+ 3PL8 )

2

dx ]W i=

12 EI {∫0

L8

9 P2x

2dx+∫

L8

L4

[4 P2x

2+2 (2Px )(PL8 )+ P2L2

64 ]dx+∫L4

L2

[P2x

2+2 (Px )( 3 PL8 )+ 9 P2 L2

64 ]dx }W i=

12 EI {∫0

L8

9 P2x

2dx+∫

L8

L4

[4 P2x

2+4 P2 Lx

8+P2L2

64 ] dx+∫L4

L2

[P2x

2+6 P2 Lx

8+

9P2L2

64 ] dx}W i=

12 EI {[3 P2 x3 ]L/8

0+[ 4 P2 x3

3+ P2Lx2

4+ P2 L2 x

64 ]L/ 4L/8

+[ P2 x3

3+3 P2 Lx2

8+ 9P2 L2 x

64 ]L/2L /4}

W i=1

2 EI {[ 3 P2L3

512 ]+[ 4 P2L3

192+ P2L3

64+P

2L3

256−4 P2 L3

1536− P2L3

256− P2L3

512 ]+[ P2L3

24+ 3 P2 L3

32+ 9P2L3

128−P2L3

192−3 P2 L3

128−9 P2L3

256 ]}

P P 2 P P P

L/8 L/8 L/8L/8L/4L/4

Ax=0Ay=3 PBy=3P

TRAMO I 0 ≤ x < L8

V=−3PM=3 Px

TRAMO II L8

≤ x < L4

V=−3P+P

M=3 Px−P(x− L8 )

M=2 Px+ PL8

TRAMO II L4

≤ x < L2

V=−3P+2P

M=3 Px−P(x− L8 )−P( x−L

4 )M=Px+ 3 PL

8

4

W i=1

2 EI {3 P2L3

512+ 49P2L3

1536+ 109 P2L3

768 }W i=

23 P2 L3

128EI

Condición:

W e=P2

(∂a+∂b+2∂c+∂d+∂e )

Ecuación de la parábola

y=A x2+Bx+C

Sistema de Ecuaciones

Donde: A=4∂

L2y B=0

Entonces:

y= 4∂

L2x2−∂

W e=P∂2 ( 7

16+ 3

4+2+ 3

4+ 7

16 )

W i=23 P2 L3

128EI

Igualamos W i=W e

23P2L3

128 EI=35

16P∂

x

y

-x

-d

(-L2 ; 0) (L2 ; 0)

(0 ; -d)0=A (−L

2 )2

+B(−L2 )+C

−∂=C

0=A ( L2 )2

+B (L2 )+C

0= AL2

4− BL

2−∂

0= AL2

4+BL

2−∂

0= AL2

4− BL

2−∂

0= AL2

4+BL

2−∂

AL2

2−2∂=0

Si: x= L4

03 L8

Es y= −3∂

4−∂

7 ∂16

5

∂=23 P L3

280 EI

Resolver la siguiente Viga.

∑ Fx=0

Ax=0

∑M=0

−By . (1 )+6( 13

.6)−5. (Ay )=0

By=92

∑ Fy=0

Ay−By=6

Donde:-Ay - By = -6 5Ay+ By = 12

Ay=32

Si:26=hx

∴ h=13x

1 m

2 Tn/m

4 m 1 m

x

92 Tn3

2 Tn

x

x

6

Momento Max. Tramo 2

Diagrama de Cortante

Diagrama de Momentos

TRAMO I 0 ≤ x < 1

V= x2

6

M=−x3

18Cuando: x=0V=0 ;M=0

x=1V=16;M=−1

18

TRAMO II 1 ≤ x < 5

V= x2

6−3

2

M=−x3

18+ 3

2( x−1 )

Cuando:

x=1V=−43

;M=−118

x=5V=83;M=−17

18

TRAMO III 5 ≤ x < 6

V= x2

6−3

2−9

2

M=−x3

18+ 3

2( x−1 )+ 9

2(x−5 )

Cuando:

x=5V=−1116

;M=−1718

x=6V=0 ;M=0

0= x2

6−3

2⟹ x2=3

26∴ x=3

Mmax=32

18+ 3

2(3−1 )∴Mmax=

32

16

83

116

43

[V]+_

7

Resolver la siguiente Carga Puntual.

Donde:

Ax=0Ay=3 PBy=3P

TRAMO I 0 ≤ x < L5

V=−3PM=3 Px

TRAMO II L5

≤ x < 2L5

V=−3P+P

M=3 Px−P(x− L5 )

M=2 Px+ PL5

TRAMO II 2L5

≤ x < L2

V=−3P+P+2P

M=3 Px−P(x− L5 )−2 P(x−2 L

5 )M=PL

178

32

118

[M]+

_

2 P

L/5 L/5 L/5 L/5

P 2 P P

L/5

8

W i=1

2 EI∫M2dx

W i=1

2 EI [2∫0L5

(3 Px )2dx+2∫L5

2L5

(2 Px+ PL5 )

2

dx+2∫2L5

L2

(PL )2dx ]W i=

1EI {∫0

L5

9P2x

2dx+∫

L5

2 L5 [4 P2

x2+( 4 P2 xL

5 )+P2 L2

25 ]dx+∫2L5

L2

[P2L

2 ] dx }W i=

1EI {3P2 x3L/5

0+ 4 P2 x3

32 L/5L/5

+ 4 P2 xL2

102 L/5L/5

+ P2L2 x25

2L/5L/5

+P2 L2 x L/22L/5}

W i=1EI {3 P2L3

125+ 32P2 L3

375−4 P2L3

375+ 8 P2 L3

125−2P2L3

125+ P2 L3

125− P2L3

125+P

2L3

2−2 P2L3

5 }W i=

1EI {3 P2L3

125+ 28P2 L3

375+ 6 P2L3

125+P

2 L3

125+ 1P2L3

10 }W i=

191 P2L3

750 EI

Condición:

W e=P2

(∂1+2∂2+2∂3+∂4 )

Ecuación de la parábola

y=A x2+Bx+C

Sistema de Ecuaciones

Donde: A=4∂

L2y B=0

Entonces:

0=A (−L2 )

2

+B(−L2 )+C

−∂=C

0=A ( L2 )2

+B (L2 )+C

0= AL2

4− BL

2−∂

0= AL2

4+BL

2−∂

0= AL2

4− BL

2−∂

0= AL2

4+BL

2−∂

AL2

2−2∂=0

Si: x= L

100

3L10

Es y= −24∂

25−∂−16∂

25

-d

(-L2 ; 0) (L2 ; 0)

d1

d2 d3

d4

x

y

-x

9

y= 4∂

L2x2−∂

W e=P∂2 (24

25+ 16

25+ 16

25+ 24

25 )W e=

8 P∂5

W i=191 P2L3

750 EI

Igualamos W i=W e

191P2 L3

750 EI=8

5P ∂

∂=191 P L3

1200 EI

Resolver la siguiente viga.

2 T/m

4.501.50

1 2

1,5m

X2 T/m

h

10

ΣFx=0 ∑Fy=0 RAx=¿ 0 RAy+RBy-6=0

RAy+RBy =6 TN

Ray= 2,5 TN= 5/2 TNΣMA=0+←

RBy (6)-4, 50(1, 50+1/3*4, 50)−1,50( 23∗1,50) =0

RBy (6) = 4, 50*3+ 1, 50 RBy = 15/6 Tn =2, 5 Tn

TRAMO I 0 ≤ X < 1, 5

2

1,5= hX

h=2 X1,50

Si x=0 {V=−3,5M=0

Si x=1.5 { V=2M=4,5

V=−72

+ X h2

=0 M=72X− X2

1,50( X

3)=0

V=−72

+ X 2

1,50 M=7

2X− X3

4,50

4,5m

X2 T/m

h

11

TRAMO II 0 ≤ X1 < 4, 5

24,5

= hX

h=2 X4,50

Diagramas de cortante.

Si x=0 {V=2,5M=0

Si x=4,5 {V=−2M=4,5

Punto de inflexion

0=52− x2

4,52 x2−22,5=0x2=11,25x=3,354m

M máx=5,59 T-m

-2Tn

-72Tn

52Tn

0( V )+

- x=3.35

V=52−X1h

2=0

M=

52X1−

X12

4,50(X 1

3)=0

V=52−

X 12

4,50 M=5

2X1−

X13

13,50

12

Diagrama del Momento

Resolver la carga puntual

M=5PX M=5PX-2P (X-L/6) M=5PX-2P(X-L/6)-P(X-L/3)

Wi = 1

2EI¿

Wi = 1

2EI¿

Wi = 1

2EI¿

Wi = 1

2EI¿

Wi =

12EI [50 P2( L3

648 )+18 P2( L3

81− L3

648 )+4 P2L( L2

18− L2

72 )+2 P2 L2

9 ( L3 −L6 )+8P2( L3

24− L3

81 )+ 16 P2 L3

( L2

8− L2

18)+ 8

9P2L2( L

2− L

3)]

Wi = 1

2EI¿

Wi = 12EI ( 199 P2L3

162 )

0

92Tn.m

5.59Tn.m

( M )+

-

13

Wi = 199 P2 L3

3 2 4 EI

y=a x2+bx+c

y=a x2+bx+c

0=a(−L2

)2

+b (−L2 )+c−δ=a (0)+b(0)+c

−δ c=c y=a x2+bx+c

0=a( L2)

2

+b( L2 )+c

aL2

4+ bL

2+c=0 a (−L

2)

2

+b (−L2 )+c=0

aL2

4−bL

2+c=0

4 δcL2 ( L2

4 )+b ( L2 )−δ c=0

2aL2

4+2c=0

b=0

a L2

2+2c=0

a=4 δcL2

y=a x2+bx+c

y=4δ c

L2 x2−δc

14

W e=12(2Pδ a+Pδ b+4 Pδ c+Pδ d+2 Pδ e)

cuando x = -L/6 Si x= -L/3

δ b=4 δcL2 (−L

6)

2

−δ c δ a=4 δcL2 (−L

3)

2

−δ c

δ b=−89

δc δ a=−59

δc

cuando x = L/6

δ d=4δ c

L2 ( L6)

2

−δ c

δ b=−89

δc

Resolver la siguiente viga.

W e=12P(−5

9δc−

89δ c−δ c−

89δ c−

59δ c)

W e=1P2

(−35δ9

)

W e=(−35Pδ18

)

W e=W i

35Pδ18

=199 P2L3

324 EI

δ c=199P L3

630EI

ΣFx=0RAy=0

ΣMA=0 +←34 ( 2

3∗1.5)−27

4 ( 23∗4.5)+RBy (4.5 )=0

RBy (4.5 )=392

RBy= 392∗4.5

=133T

ΣFy=0

RAy+RBy=34+ 23

4

RAy+RBy=152T

RAy=152

−133

RAy=196T

1 Tn/m

3 Tn/m

RAy

1.50 4.50

1 2

15

TRAMO I 0 <= X < 1.5

TRAMO II 1.5 <= X < 6

11.5

= h1.5−X

h=(1.5−X )

1.5

V=1+ 1.5−X

1.52

∗X=

1.5+1.5−X1.52

∗X=

3−X1.52

∗X=3−X

3∗X

M=−[( 1.5−X1.5

∗X ) X2 +X (1−1.5−X

1.5 )2

∗( 23X )]

M=−[ 1.5−X3

∗X 2+

X ( 1.5−1.5−X1.5 )2

∗2

3X ]

Si x=0 {V=0M=0

Si x=1.5 { V=3/ 4M=−3/ 4

V= 34−19

6+

2(X−1.5)2

6=−29

12+(X−1.5)2

3

V=−34

(X−0.5 )+196

(X−1.5 )−(X−1.5)3

9

Si x=1.5 {V=−29 /12M=−3/4

Si x=6 {V=13 /3M=0

34.5

= hX−1.5

h=2(X−1.5)

3

PUNTO DE INFLEXION

0=−2912

+(X−1.5)2

2(X−1.5)2

2=29

12

(X−1.5)2=5812

16

Diagramas de momento y cortante realizados en Excel.

Diagrama de Cortantes

X CORTANTE MOMENTO0 0,00 0,00

0,5 0,42 -0,111,5 0,75 -0,751,5 -2,42 -0,752 -2,33 0,443 -1,67 2,50

3,7 -0,80 3,384 -0,33 3,565 1,67 2,946 4,33 0,00

TRAMO I 0 <= X < 1.5

TRAMO II 1.5 <= X < 6

PUNTO DE INFLEXION

0=−2912

+(X−1.5)2

2(X−1.5)2

2=29

12

(X−1.5)2=5812

0

133 Tn

34Tn

2912Tn

( V )+

-

17

Diagrama de momentos

Resolver la carga puntual

34Tn.m

3.59Tn.m

( M )+

-

Wi= 12 EI∫M 2dx

We=12

(4 Pδ 1+3 Pδ2+3 Pδ3+4 Pδ 4 )M=7 Px

M=7 Px−4 P(x− L6 )=7 Px−4 Px+ 2

3PL=3 pX+ 2

3PL

18

Wi= 12 EI∫M 2dx

We=12

(4 Pδ 1+3 Pδ2+3 Pδ3+4 Pδ 4 )M=7 Px

M=7 Px−4 P(x− L6 )=7 Px−4 Px+ 2

3PL=3 pX+ 2

3PL

Wi= 1EI

¿

Wi= 1EI

[(49 P2 X3

3 )0

L6 +(9 P2 X

3

3+4 P2L

X2

3+ 4

9P2L2 X )L

6

L3 +( 25

9P2 L2 X)L

3

L2 ]

Wi= 1EI

[(49 P2 X3

3 )0

L6 +(3 P2 X3+2P2 L X2+ 4

9P2L2 X)L

6

L3 +(25

9P2L2 X )L

3

L2 ]

Wi= 1EI

[ 493P2( L6 )

3

+3 P2 X ( L3 )3

+2P2L X ( L3 )2

+ 49P2L2( L3 )−3 P2 X ( L6 )

3

−2 P2L X ( L6 )2

−49P2 L2(L

6)+ 25

9P2 L2 L

2−25

9P2L2 L

3]

Wi= 1EI

¿

Wi= 1EI ( 71

81P2L3)= 71

81EIP2L3

19

We=Wi

y= δ9−δ

12

(4 Pδ1+3Pδ 2+3 Pδ3+4 Pδ4 )= 7181EI

P2L3

12 (4 P

59δ+3P

89δ+3 P

89δ+4 P

59δ)= 71

81EIP2L3

Pδ2 ( 20

9+ 24

9+24

9+ 20

9 )= 7181EI

P2 L3

Pδ2 ( 88

9 )= 7181EI

P2 L3

Pδ (44 )= 719 EI

P2L3

δ= 71 P2L3

9∗44 EI=71 P2L3

396 EI

y (−L6 )( L6 )=4 δ

L2 ( L6)

2

−δ

y= δ9−δ

y=−89

δ (δ2 , δ 3)

y= 4δ

L2X2−δ

y (−L3 )( L3 )=4 δ

L2 ( L3)

2

−δ

y= 4δ9

−δ

y=−59

δ(δ1 , δ 4)

y=a x2+bx+c

(−L2

,0)

0=a(−L2

)2

+b (−L2

)+c

0=aL2

4−bL

2+c

(0 ,−δ)−δ=CC=−δ

( L2,0)

0=a( L2)

2

+b (L2)+c

0=aL2

4+ bL

2+c

0=aL2

4−bL

2+c

0=aL2

4+ bL

2+c

0=aL2

2+0+2c

a L2

4=−2c

a=4 δ

L2

( 4 δ

L2 ) L4 +b (L2 )=δ

δ+b=δb=0