resistencia de materiais. tema 9. cálculo de movementos...

Resistencia de Materiais. Tema 9. Cálculo de movementos en estruturas de barras

ARTURO NORBERTO FONTÁN PÉREZ

Fotografía. Sunniberg (Christian Menn, Klosters, Suiza, 1999).

Van principal: 140 m.

Contido. Tema 9. Cálculo de movementos en estruturas de barras

2

1. Introdución. 2. Integración por tramos da ecuación diferencial asociada á deformación da barra. 3. Cálculo de movementos por integración das deformacións. Fórmulas de Bresse.

Resistencia de materiais. Tema 9 ETS Enxeñeiros de Camiños, Canais e Portos

Fotografía. Ganter (Christian Menn,

Valais, Suiza, 1980). Van principal: 174 m.

9.1. Introdución

3

Unha barra solicitada por unhas cargas exteriores experimenta uns movementos, que se poden expresar co vector de movementos:

As ecuacións diferenciais que relacionan os esforzos interiores cos movementos son: En xeral, na maioría dos elementos barra solicitados a flexión simple, pode comprobarse que os movementos

provocados polo esforzo cortante son desprezables comparados cos movementos provocados polo momento flector.

Ademais, tamén se pode demostrar que os movementos producidos pola flexión e a torsión son da mesma orde de magnitude, mentres que o correspondente ao esforzo axil é moi inferior. Polo que se pode prescindir del en boa parte das situacións e o cálculo dos movementos nunha barra soe facerse considerando só os producidos por momentos torsores e flectores.

Sen embargo, non sempre se pode prescindir dos movementos por esforzo axil e cortante.


x y zu v w ϕ ϕ ϕ = u

Esforzo axil N

Momentos flectores My, Mz

Esforzos cortantes Vy, Vz

Torsión uniforme Mx

du Ndx E A

=⋅

x x

T

d Mdx G Iϕ

=⋅

2y y

2y

d Md wdx dx E Iϕ

− = = −⋅

2z z

2z

d d v Mdx dx E Iϕ

= =⋅

2y z z

2z z

d d w dV pdx dx G A dx G Aϕ

− = = = −⋅ ⋅ ⋅

2y yz

2y y

dV pd d vdx dx G A dx G Aϕ

= = − =⋅ ⋅ ⋅


4



Fotografía. Chandoline (Christian Menn, Sion,

Suiza, 1989). Van principal: 140 m.

9.2. Integración por tramos da ecuación diferencial asociada á deformación da barra

5


Obtención das leis de esforzos

Integración por tramos das ecuacións diferenciais

Imposición das condicións de contorno

Tipo de apoio Condicións de contorno

Apoio simple

Frecha nula

Momento flector nulo

Articulación

Frecha nula

Momento flector nulo

Empotramento

Frecha nula

Xiro nulo

Extremo libre Forzas exteriores na sección

w 0= v 0=

2

2d w 0dx

=2

2d v 0dx

=

w 0= v 0=

2

2d w 0dx

=2

2d v 0dx

=

w 0= v 0=

dw 0dx

=dv 0dx

=

2yA

2y

Md wdx E I

= −⋅

2zA

2z

d v Mdx E I

= −⋅

3zA

3y

d w Vdx E I

= −⋅

3yA

3z

Vd vdx E I

= −⋅

PPzA

MyA

xAA


6

C.


Tipo de apoio Condicións de contorno (continuación)

Articulación interior

Igualdade de frecha

Momento nulo

Equilibrio de forzas na rebanada

Apoios interiores

Frecha nula

Igualdade de xiros

Igualdade de momentos

Sección interior con cargas exteriores illadas

Igualdade de frechas

Igualdade de xiros

Equilibrio de momentos

Equilibrio de forzas

d fw w= d fv v=

22fd

2 2

d wd w 0dx dx

= =22

fd2 2

d vd v 0dx dx

= =

33fd

y3 3

d vd v P 0dx dx

− + =33

fdz3 3

d wd w P 0dx dx

− + =

d fw w 0= =

fd dwdwdx dx

=

22fd

2 2

d wd wdx dx

=

d fv v 0= =

fd dvdvdx dx

=

22fd

2 2

d vd vdx dx

=

d fw w= d fv v=

fd dwdwdx dx

= fd dvdvdx dx

=

33fd

z3 3

d wd w P 0dx dx

− + =33

fdy3 3

d vd v P 0dx dx

− + =

22f yAd

2 2y

d w Md w 0dx dx E I

− + + =⋅

22fd zA

2 2z

d vd v M 0dx dx E I

− + + =⋅

d

Pz

f

d f

d f

d f

PzA

MyA

A


7

Exemplo 1. Obtención de movementos. Viga en voladizo con carga puntual no extremo. 1. Obter a lei de momentos flectores:

2. Integrar por tramos a ecuación diferencial:

3. Impoñer as condicións de contorno:

4. Os movementos resultan:


( )yM P L x= ⋅ −

( ) ( ) ( )2 32y

1 1 22y y y y

M P L x P L x P L xd w dw 1 1c w c x cdx E I E I dx 2 E I 6 E I

⋅ − ⋅ − ⋅ −= − = − ⇒ = + ⋅ ⇒ = − + ⋅ + ⋅

⋅ ⋅ ⋅ ⋅

3 2

x 0 2 1

2 3

1 2x 0

P L P Lw 0 c 0 c6 2dw P L P L0 c 0 cdx 2 6

=

=

⋅ ⋅= ⇒ − + = = − ⇒ ⋅ ⋅ = ⇒ + = =

( )3 2 3

y

L x L L Pw x6 2 6 E I

−= − − ⋅ + ⋅

⋅ 3

máx x Ly

P Lw w3 E I=

⋅= = −

⋅ ⋅

LP

P·LA Bx

P


8

Exemplo 2. Obtención de movementos. Viga en voladizo con carga distribuída uniforme. 1. Obter a lei de momentos flectores:





( )2

yq L x

M2

⋅ −=

( ) ( ) ( )2 3 42y

1 1 22y y y y

M q L x q L x q L xd w dw 1 1c w c x cdx E I 2 E I dx 6 E I 24 E I

⋅ − ⋅ − ⋅ −= − = − ⇒ = + ⋅ ⇒ = − + ⋅ + ⋅

⋅ ⋅ ⋅ ⋅ ⋅

4 3

x 0 2 1

3 4

1 2x 0

q L q Lw 0 c 0 c24 6

dw q L q L0 c 0 cdx 6 24

=

=

⋅ ⋅= ⇒ − + = = − ⇒ ⋅ ⋅ = ⇒ + = =

( )4 3 4

y

L x L L qw x24 6 24 E I

−= − − ⋅ + ⋅

⋅ 4

máx x Ly

q Lw w8 E I=

⋅= = −

⋅ ⋅

Lq·L

q·L²2

A B

x


9

Exemplo 3. Obtención de movementos. Viga biapoiada con carga puntual central. 1. Obter a lei de momentos flectores:





( )

Tramo AB:

Tramo BC :

y

y

L P x0 x M2 2

P L xL x L M2 2

⋅≤ ≤ → = −

⋅ −≤ ≤ → = −

( ) ( )

( ) ( )

;AB BC 2 3 4x 0 x L 22 2

1 3AB BC1 3

x L 2 x L 2 3

3 3 41 3

AB BC 2 4x L 2 x L 2

w w 0 c 0 c L c 0P Lc cdw dw P L P Lc c 16

dx dx 16 16P LcP L c L P L c L 16w w c c

96 2 96 2

= =

= =

= =

= = ⇒ = ⋅ + = ⋅

= − = −⋅ ⋅ = ⇒ + = − + ⇒ ⋅ = −⋅ ⋅ ⋅ ⋅ = ⇒ + + = + +

( )33 2 2 3 3

AB BC máx x L 2y y y

L xx L x P L x L P P Lw w w w12 16 E I 12 16 16 E I 48 E I=

− ⋅ ⋅ ⋅= − ⋅ = + − ⋅ = = − ⋅ ⋅ ⋅ ⋅

( ) ( ) ( )

2 2 3yAB AB

1 AB 1 22y y y y

2 32BC BC

3 BC 3 42y y y

Md w P x dw P x 1 P x 1c w c x cdx E I 2 E I dx 4 E I 12 E I

P L x P L x P L xd w dw 1 1c w c x cdx 2 E I dx 4 E I 12 E I

⋅ ⋅ ⋅= − = ⇒ = + ⋅ ⇒ = + ⋅ + ⋅ ⋅ ⋅ ⋅ ⋅ ⋅

⋅ − ⋅ − ⋅ −= ⇒ = − + ⋅ ⇒ = + ⋅ + ⋅

⋅ ⋅ ⋅ ⋅

P

P2

A C

B

P2

L2

xL2


10

Exemplo 4. Obtención de movementos. Viga biapoiada con carga distribuída. 1. Obter a lei de momentos flectores:





2

yq x q L xM

2 2⋅ ⋅ ⋅

= −

3x 0 2

14 4

x L 1 22

w 0 c 0 q Lc24q L q Lw 0 c L c 0 c 024 12

=

=

= ⇒ = ⋅= ⇒ ⋅ ⋅

= ⇒ − + ⋅ + = =

4 3 3

y

x L x L x qw24 12 24 E I

⋅ ⋅= − + ⋅ ⋅

2 2 3 2y

12y y y

4 3

1 2y

Md w q x q L x 1 dw q x q L x 1cdx E I 2 2 E I dx 6 4 E I

q x q L x 1w c x c24 12 E I

⋅ ⋅ ⋅ ⋅ ⋅ ⋅= − = − ⋅ ⇒ = − + ⋅ ⇒ ⋅ ⋅ ⋅

⋅ ⋅ ⋅⇒ = − + ⋅ + ⋅ ⋅

qL2 qL

2Lx

q

4

máx x L 2y

5 q Lw w384 E I=

⋅ ⋅= = −

⋅ ⋅


11

Coas expresións dos movementos en estruturas isostáticas poden resolverse estruturas hiperestáticas. Exemplo 5:


, ,B q B Rw w 0+ =

,

, ,

,

4

B q 4 3y

B q B R3y y

B Ry

5 q Lw384 E I 5 q L R L 5 q Lw w 0 0 R

384 E I 48 E I 8R Lw48 E I

⋅ ⋅= − ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅⇒ + = ⇒ − + = ⇒ = ⋅ ⋅ ⋅ ⋅⋅ = ⋅ ⋅

L2

L2

VA VB VC

A B C

q

L2

L2

VA VC

A B C

q

R

=+


12



Fotografía. Leonard P. Zakim Bunker Hill (Christian Menn,

Boston, EEUU, 2005). Van principal: 227 m.

9.3. Cálculo de movementos por integración das deformacións. Fórmulas de Bresse.

13

Nunha estrutura como a da figura, os movementos na barra 1-2 dependen das cargas, do material (E, G) e dos parámetros mecánicos da barra (A, Az e Iy), mentres que na barra 2-3 só dependen dos movementos da sección 2 e da xeometría da barra porque sobre ela non actúan directamente cargas exteriores. Os primeiros denomínanse movementos de sólido elástico e os segundos movementos de sólido ríxido.

A expresión xeral dos movementos de sólido ríxido nunha sección calquera da barra 2-3 de coordenadas (X,Z) resulta:

Engadindo cargas exteriores na barra 2-3, os movementos da barra 1-2 cambian e aparecen movementos de

sólido elástico na barra 2-3. Para obter os movementos de sólido elástico considérese unha sección de coordenadas (X,Z) e unha rebanada de lonxitude dx:


( )

( )

sen

cos

3 2 3 23 2 2 y 3 2

23 2 y 23

3 2 3 23 2 2 y 3 2

23 2 y 23

Z Z u u u u Z ZL L

X X w w w w X XL L

β ϕϕ

β ϕϕ

− −= = ⇒ = + ⋅ −

⋅

− −= = ⇒ = − ⋅ −

− ⋅

( ) ( )y 2 y 2 2 y 2 2 2 y 2u u Z Z w w X Xϕ ϕ ϕ ϕ= = + ⋅ − = − ⋅ −

''

'

y zy sz

y z

MN Vd dx d dx dx dxE A E I G A

δ ϕ γ= ⋅ = ⋅ ⋅ = ⋅⋅ ⋅ ⋅

Nós ríxidos! Non hai xiro relativo.

1 2

P

3

Z

X

2

3u

wβ

φy2

β

3

3

w2

φy2

1

P

Z

XP2

1

(X,Z)

dxdδ

dx

dφy

dx

γsz'

γ ·dxsz'


14

Os movementos de sólido elástico desta rebanada producen movementos de sólido ríxido no resto da barra: Integrando as expresións anteriores ao longo da barra obtéñense os movementos como sólido elástico da

sección 3. Os movementos totais son estes máis os movementos como sólido ríxido:


( )

( )

( )

( )

''

'

''

'

yy3

yy3 y

y z3 y 3 sz 3 3

y z

y z3 y 3 sz3 3

y z

Md dx

E Id 0 d 0MdX dZ N Vdu d d Z Z dx du dX dx Z Z dZ

dx dx E A E I G AdZ dX MN Vdw d d X X dx dw dZ dx X X dXdx dx E A E I G A

ϕϕ ϕ

δ ϕ γ

δ ϕ γ

= ⋅

⋅= + + = ⋅ + ⋅ − − ⋅ ⋅ ⇒ = ⋅ + ⋅ ⋅ − − ⋅ ⋅ ⋅ ⋅ = ⋅ − ⋅ − + ⋅ ⋅ = ⋅ − ⋅ ⋅ − + ⋅ ⋅ ⋅ ⋅

( ) ( )

( ) ( )

'

'

'

'

3y

y3 y2y2

3 3 3y z

3 2 y2 3 2 3y z2 2 2

3 3 3y z

3 2 y2 3 2 3y z2 2 2

Mdx

E I

MN Vu u Z Z dX Z Z dx dZE A E I G A

MN Vw w X X dZ X X dx dZE A E I G A

ϕ ϕ

ϕ

ϕ

= + ⋅⋅

= + ⋅ − + ⋅ + ⋅ − ⋅ − ⋅⋅ ⋅ ⋅

= − ⋅ − + ⋅ − ⋅ − ⋅ + ⋅⋅ ⋅ ⋅

∫

∫ ∫ ∫

∫ ∫ ∫

Fórmulas de Bresse

De aplicación en estruturas de nós ríxidos

2

3dδ dwdu

2

3 du-dw

dφ y

2

3

-dudwγ ·dx

sz'


15

Pode demostrarse que as contribucións do esforzo axil e do esforzo cortante son insignificantes respecto ás producidas pola flexión, polo que soe utilizarse unha formulación reducida. Os movementos dunha sección xenérica calquera S (Xs, Zs) en función dos movementos dunha sección 1 poden expresarse como:

Exemplo 1a. Cálculo dos movementos no extremo B da estrutura da figura:


( ) ( )

( ) ( )

sy

ys y1y1

sy

s 1 y1 s 1 sy1

sy

s 1 y1 s 1 sy1

Mdx

E I

Mu u Z Z Z Z dx

E I

Mw w X X X X dx

E I

ϕ ϕ

ϕ

ϕ

= + ⋅⋅

= + ⋅ − + ⋅ − ⋅⋅

= − ⋅ − − ⋅ − ⋅⋅

∫

∫

∫

Ollo! Non confundir x, X, Z!

yA A A0 u 0 w 0ϕ = = = ( ) ( )yM x P L x= ⋅ −

( )

( )

( ) ( )

( ) ( )

( ) ( ) ( ) ( )

s B L 2y

ys yBy y y yA A 0

s By

s s B By yA A

s B L 3y

s s B By y y yA A 0

M P L x P P Ldx dx L x dxE I E I E I 2 E I

M P L xu Z Z dx u Z Z dx 0

E I E I

M P L x P L x P Lw X X dx w X X dx L x dxE I E I E I 3 E I

ϕ ϕ ⋅ − ⋅

= ⋅ = ⋅ = ⋅ − ⋅ = ⋅ ⋅ ⋅ ⋅ ⋅

⋅ −= ⋅ − ⋅ ⇒ = ⋅ − ⋅ = ⋅ ⋅ ⋅ − ⋅ − ⋅= − ⋅ − ⋅ = − ⋅ − ⋅ = − ⋅ − ⋅ = −

⋅ ⋅ ⋅ ⋅ ⋅

∫ ∫ ∫

∫ ∫

∫ ∫ ∫

x X dx dX= ⇒ =L

P

P·LA Bx

P


16

Exemplo 1b. Cálculo dos movementos na sección xenérica S da estrutura da figura: Exemplo 2. Cálculo dos movementos no extremo B dunha viga en voladizo con carga distruída:


LP

P·LA Bx

P

x

S

s

yA A A0 u 0 w 0ϕ = = = ( ) ( )yM x P L x= ⋅ −x X dx dX= ⇒ =

( )

( )

( )

( ) ( )

( ) ( )

s

s

xsy s

ys yS sy y yA A

s Sy

s s S sy yA A

s x 2y s s

s s S syA y yA

M P L x P xdx dx x LE I E I E I 2

M P L xu Z Z dx u Z Z dx 0

E I E I

M P L x P x xw X X dx w X X dx LE I E I E I 2 3

ϕ ϕ ⋅ − = ⋅ = ⋅ = ⋅ ⋅ − ⋅ ⋅ ⋅ ⋅ − = ⋅ − ⋅ ⇒ = ⋅ − ⋅ = ⋅ ⋅

⋅ − = − ⋅ − ⋅ = − ⋅ − ⋅ = − ⋅ ⋅ − ⋅ ⋅ ⋅

∫ ∫

∫ ∫

∫ ∫

Lq·L

q·L²2

A B

xyA A A0 u 0 w 0ϕ = = = ( ) ( )2

yL x

M x q2−

= ⋅x X dx dX= ⇒ =

( )

( ) ( )

( ) ( )

2L 3

yBy y0

2L

B sy0

2L 4

By y0

q L x q Ldx2 E I 6 E I

q L xu Z Z dx 0

2 E I

q L x q Lw L x dx2 E I 8 E I

ϕ ⋅ − ⋅

= ⋅ =⋅ ⋅ ⋅ ⋅

⋅ − = ⋅ − ⋅ = ⋅ ⋅

⋅ − ⋅ = − ⋅ − ⋅ = − ⋅ ⋅ ⋅ ⋅

∫

∫

∫


17

Exemplo 3. Cálculo dos movementos máximos: Exemplo 4. Cálculo dos movementos máximos:


yB A A C0 u 0 w w 0ϕ = = = =

x X dx dX= ⇒ =

( ) ( )

L 2B 2y

yB yA yA yAy y yA 0

B L 2 3y

B A yA B A B B yAyA y y0

M P x P Ldx 0 dxE I 2 E I 16 E I

M L P x L P Lw w X X X X dx w 0 x dxE I 2 2 E I 2 48 E I

ϕ ϕ ϕ ϕ

ϕ ϕ

− ⋅ ⋅= + ⋅ = + ⋅ ⇒ =⋅ ⋅ ⋅ ⋅ ⋅ ⇒

− ⋅ ⋅ = − ⋅ − − ⋅ − ⋅ = − ⋅ − ⋅ − ⋅ = − ⋅ ⋅ ⋅ ⋅ ⋅

∫ ∫

∫ ∫

( )y

y

L P x0 x M2 2

P L xL x L M2 2

⋅≤ ≤ → = −

⋅ −≤ ≤ → = −

( )2

yq x q L x q xM x L

2 2 2⋅ ⋅ ⋅ ⋅

= − = ⋅ −yB A A C0 u 0 w w 0ϕ = = = =

x X dx dX= ⇒ =

( ) ( )

( )

( )

L 2B 3y

yB yA yA yAy y yA 0

B L 2 4y

B A yA B A B B yAyA y y0

M q x q Ldx 0 x L dxE I 2 E I 24 E I

M q x x LL L 5 q Lw w X X X X dx w 0 x dxE I 2 2 E I 2 384 E I

ϕ ϕ ϕ ϕ

ϕ ϕ

⋅ ⋅= + ⋅ = + ⋅ − ⋅ ⇒ =⋅ ⋅ ⋅ ⋅ ⋅ ⇒

⋅ ⋅ − ⋅ ⋅ = − ⋅ − − ⋅ − ⋅ = − ⋅ − ⋅ − ⋅ = − ⋅ ⋅ ⋅ ⋅ ⋅

∫ ∫

∫ ∫

qL2 qL

2Lx

CBA

P

P2

A C

B

P2

L2

xL2


18

Exemplo 5. Resolución dunha estrutura hiperestática:


P

A C

B

L2

L2

P

A C

B

R

L2

L2

, ,C P C Rw w 0+ =+=yA A0 w 0ϕ = =

Estrutura real Estrutura isostática equivalente

( ) ( ) ( ), , , ,

, , ,

L 2C 3y

C P A P yA P C A C C Py y yA 0

3 3 3

C R C P C Ry y y

LP xM 5 P L2w w X X X X dx w L x dxE I E I 48 E I

R L 5 P L R L 5 Pw w w 0 0 R3 E I 48 E I 3 E I 16

ϕ

⋅ − ⋅ ⋅ = − ⋅ − − ⋅ − ⋅ ⇒ = − ⋅ − ⋅ = −⋅ ⋅ ⋅ ⋅

⋅ ⋅ ⋅ ⋅ ⋅= + = ⇒ − + = ⇒ =

⋅ ⋅ ⋅ ⋅ ⋅ ⋅

∫ ∫

-

+

M

V+

-

3PL16

11P16

5P16

5PL32

yA yA

A A

B

L 3 P LM R L P 0 M2 16

11 PV R P 0 V16

L 5 P LM R2 32

⋅ ⋅− ⋅ + ⋅ = ⇒ = −

⋅+ − = ⇒ =

⋅ ⋅= − ⋅ = −

, , yAA P A M 0ϕ ϕ+ =

P

A C

B

P

A C

BA

MyA

=

+

Outra forma:

resistencia de materiais. tema 9. cálculo de movementos...

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