s8 PPD1 Teorie Polin Navier

7/24/2019 s8 PPD1 Teorie Polin Navier

1/27

PLCI PLANE

DREPTUNGHIULARE

1


2/27

ECUAIA FUNDAMENTALa plcilor plane dreptunghiulare

Ipoteze

x

y

z

h

x

yl

l

p x,y( )= ct

yx lla

a

h

,min

20

1

80

1

5

1max h

w

Plci plane subiri cu deformaii mici


3/27

Ipoteze

Ipoteza lui Kirchhoff:O dreapt normal la planul

median nainte de deformare rmne dreapt inormal la planul median dup deformare

Planul median este inextensibil n planul su

zeste neglijabil n raport cu x i y


4/27

Aspectul static

xz

x

x

yz

dx

yd

h/2

h/2

zdz

1

1

y

xy

yz

yx

Txz

MxxyM

yzT

yM

Myx

x

y

z


5/27

Aspectul static

Txz

Mx

xyMyzT

yM

Myx

x

y

z


6/27

Aspectul geometric

y

w

x

w

y

x

x, y

z

h

w

u, v

x

y

,

z

zv

zu

y

x


7/27

Ecuaia Sophie Germain - Lagrange

D

yxpyxw

,,22

2

2

2

22

yx

D

yxp

y

w

yx

w

x

w ,2

4

4

22

4

4

4

ecuaie diferenial liniar de ordinul IV,neomogen, cu coeficieni constani

23

112

EhD

rigiditatea cilindricla ncovoiere a plcii


8/27

Eforturi secionale

2

2

2

2

y

w

x

wDMx

2

2

2

2

xw

ywDMy

wx

Dy

M

x

MT

yxxxz

2

wy

D

y

M

x

MT

yxyyz

2

momente ncovoietoare

momente de torsiune yx

wDMM yxxy

2

1

fore tietoare


9/27

CONDIII DE CONTUR

x

y

z

h

b

aO A

B C

latur ncastrat

latur simplurezemat

latur ncastratelastic

latur liber


10/27

Condiii de conturx

y

z

h

b

aO A

B C

0w 0

y

wy 02

2

x

wx

002

yx

w

x y

2

2

y

wDMx

2

2

y

wDMy

0xyM

yx MM

3

3

y

wD

y

MT

yyz

Latura ncastrat OA (y = 0)


11/27

Condiii de conturx

y

z

h

b

aO A

B C

0w 0xM 022

2

2

y

w

x

w

02

2

ywy

02

2

x

w

Latura simplu rezemat OB (x= 0)

Fora tietoare generalizat

010022

yx

wDM

yx

w

y xyxyx

2

3

3

3* 2

yx

w

x

wD

y

MTT

xy

xzxz

laturiilunguln.ctx

y

z

xyM dy

dyxyM

dy

(M + ...)xy

(M + )xy


12/27

Condiii de conturx

y

z

h

b

aO A

B C

Latura cu rezemare elastic AC (x = a)

y

xzR

Mx

Rxz

xM

-solicitarea la ncovoiere a grinzii n planul yOz :

xzRxq

EI

xq

y

w

;

4

4

gr

xz

EI

R

y

w

4

4

2

3

3

3

4

4

2yx

w

x

w

EI

D

y

w

gr

-solicitarea la torsiune a grinzii:

y

xtt

t dyyMMGI

M0

; xt

t

t

MGIdy

dM

GIy

11

2

2

2

2

2

3

y

w

x

w

GI

D

yx

w

t

yx

w

y

x

2


13/27

Condiii de conturx

y

z

h

b

aO A

B C

Latura liber BC (y = b)

yx

w

y

wD

x

MTR

yx

yzyz 2

3

3

3

2

Fora tietoare generalizat

0,0 yzy RM sau 00 , QRMM yzy

DM

xw

yw 0

2

2

2

2

/0

D

Q

yx

w

y

w 02

3

3

3

/02


14/27

- plci libere de legturi, ncrcate pe contur cu momente ncovoietoarei/sau de torsiune distribuite uniform

0,22 yxw feydxcybxyaxyxw 22,

REZOLVAREA PLCILOR PLANEDREPTUNGHIULARE

Soluii exacte cu polinoame algebrice:

caDy

w

x

wDMx

2

2

2

2

2

acDx

w

y

wDMy

2

2

2

2

2

bDyx

wDMxy

11

2

1M

2M12M

21M


15/27

0M

0M

Exemple:O

b /2

y

b /2

x

z

a /2 /2a

placa supus la ncovoiere

0; 12021 MMMM

012

0

bD

Mca

feydxyx

D

Mw

220

12

.ctyx

paraboloid de revoluie

d = e = f = 0 dac se consider c n punctul O (lax=0 iy=0)

0w

0x

w

0

y

w

220

12yx

D

Mw


16/27

placa supus la torsiune

01221 ;0 MMMM

1

0

0

D

Mb

ca

xy

D

Mw

1

0 .ctxy

0M

0M

M02 2M0

M2 02M0

paraboloid hiperbolic

d = e = f = 0


17/27

serii duble trigonometriceNavier:

plciplane dreptunghiulare simplu rezemate pe contur; origineasistemului de axe trebuie sfie ntr-un colal plcii

depinde dencrcare

ncrcaredistribuitsinusoidal ncrcaredistribuituniform, localsau pe toat suprafaa plcii ncrcare distribuit liniar forconcentrat

ybn

xa

m

wyxwn

mnm

11 sinsin,

mnw ybn

xa

mpyxp

nmn

m

11

sinsin,

ydxdybn

xa

m

yxpabp

ba

mn

sinsin,

4

00

REZOLVAREA PLCILOR PLANEDREPTUNGHIULARE

Soluii cu serii Fourier = exacte pentru un numr infinit de termeni


18/27

serii simple trigonometrice: Lvy plci plane dreptunghiulare simplu rezemate pe dou laturi paralele,

orice tip de rezemare pe celelalte laturi; axa pe laturile simplu

rezemate trebuie s fie ax de simetrie

yxwyxwyxw p ,,, 0 0,0

22 yxw

D

yxpyxwp

,,22

02sin,4

44

2

22*

10

mm

IVm

mm Y

amY

amYx

amyYyxw

a

myyDyyCyByAyY mmmmmmmmmm

,shchshch)(

(*)

(**)

analog grinda-perete

dac

D

xp

dx

wdxpp

p 4

4**dezvoltare n serie simpl trigonometric

Obs. Seriile simple trigonometrice sunt mult mai rapid convergente dectseriile duble trigonometrice.


19/27

SOLUII APROXIMATIVE

metode variaionale: Ritz - Galerkinteorema minimului energiei poteniale totale Galerkinprincipiul lucrului mecanic virtual

metode numerice: metoda diferenelor finite metoda elementului finit metoda fiilor finite


20/27

ax

y

b

z

A B

C D

p( )x,yAPLICAREASOLUIEINAVIER

,3,2,1,;;sinsin,11

nminecunoscuicoeficienwyb

nx

a

mwyxw mn

nmn

m

Verificarea condiiilor de margine:AB, CD(y= 0,y= b)

0

0sinsin

00

12

22

12

2

0

2

2

2

2

y

nmn

m

y

y

M

yb

nx

a

m

b

nw

y

w

x

w

y

wDM

Mw

AC,BD(x= 0,x= a)

0

0sinsin

00

12

22

12

2

0

2

2

2

2

x

nmn

m

x

x

M

yb

nx

a

m

a

mw

x

w

yw

xwDM

Mw

D

yxpyxw

,,22

2

3

112;

Eh

D

2224


21/27

yb

nx

a

m

a

mw

x

w

nmn

m

1

2

2

22

14

4

sinsin

yb

nx

a

m

b

n

a

mw

yx

w

nmn

m

12

22

2

22

122

4

sinsin

yb

nx

a

m

b

nw

y

w

nmn

m

1

2

2

22

14

4

sinsin

y

b

nx

a

m

b

n

a

mww

nmn

m

1

2

2

22

2

22

1

22 sinsin

D

yxp ,

Dezvoltarea ncrcrii n serie dubl trigonometric:

yb

nx

a

mpyxp

nmn

m

11

sinsin,

D

p

b

n

a

mw mnmn

2

2

2

2

24 dxdy

b

nx

a

myxp

abp

ba

mn

sinsin,

4

00

I

bamn

mn ydxdyb

nx

a

myxp

bn

amDab

bn

am

p

Dw

sinsin,

41

002

2

2

2

2

4

2

2

2

2

24

a


22/27

Exemplu

ax

y

b

x0

0y

c c

dd

z

pctyxp ,

dydyy

cxcxx

00

00

,

,

21sinsin0

0

0

0

IpIdxdyb

nx

a

mpI

dy

dy

cx

cx

ca

mx

a

m

m

a

a

m

xa

m

xdxa

mI

cx

cx

cx

cx

sinsin2cos

sin 01

0

0

0

0

dbny

bn

nb

b

n

y

b

n

ydybnI

dy

dy

dy

dy

sinsin2

cos

sin 02

0

0

0

0

db

ny

b

nc

a

mx

a

m

b

n

a

mDmn

pwmn

sinsinsinsin16

002

2

2

2

26

b


23/27

Caz particular: ncrcare distribuit uniform pe toat placa2

;2

00

bdy

acx

,5,3,1,;16

2

2

2

2

26

nm

b

n

a

mDmn

p

,5,3,1,;sinsin

16,

12

2

2

2

21

6

nm

b

n

a

mmn

yb

nx

a

m

D

pyxw

nm

,5,3,1,;sinsin16

12

2

2

2

2

2

2

2

2

142

2

2

2

nmyb

nx

a

m

b

n

a

mmn

bn

amp

y

w

x

wDM

nm

x

,5,3,1,;sinsin161

2

2

2

2

2

2

2

2

2

142

2

2

2

nmy

bnx

am

b

n

a

mmn

a

m

b

n

pxw

ywDM

nm

y

,5,3,1,;coscos

1611

1

2

2

2

2

2

1

4

2

nm

bn

am

yb

nx

a

m

ab

p

yx

wDM

nm

xy

2sin

2sin

16 222

2

2

2

26

nm

b

n

a

mDmn

pwmn


24/27

Plac ptrat din oel (a = b; = 0,3):

,5,3,1,;

sinsin16

,1

2221

6

4

nmnmmn

yb

nx

a

m

D

payxw

nm

,5,3,1,;sinsin

16

1222

22

14

2

nmyb

nx

a

m

nmmn

nmpaMM

nm

yx

,5,3,1,;

coscos16

1 1 22214

2

nmnm

yb

nx

a

m

pa

M nmxy


25/27

n centrul plcii, la x=y= a/2:

,5,3,1,;16

116

4

max

nmtD

paw

n

mn

m

2221

2

222

12sin

2sin

nmmnnmmn

nm

t

nm

mn

D

pa

ttttttttttttttttD

pat

D

pa

D

pattttttttt

D

pat

D

pa

D

patttt

D

pat

D

pa

Dpat

D

patD

pa

w

nmn

m

n

mn

m

nmn

m

nmn

m

4

777557733771175553355115333113116

47

1

7

16

4

4

5553355115333113116

45

1

5

1

6

4

4

333113116

43

1

3

16

4

4

116

41

1

1

16

4

max

004045931,0

1616

0040647,01616

004055,01616

004161,01616

,5,3,1,;16

114

2

max,max,

nmt

paMM

n

mn

m

yx

222221

2

222

22 1

2sin

2sin

nmmn

nmnm

nmmn

nmt

nm

mn

2

2

2

max

0482,0

04692,0

05338,0

pa

pa

pa

M

0,5,3,1,;2cos

2

cos

1611

2221

4

2

xy

nm

xy Mnmnm

nm

paM


26/27

La colurile plcii:

A(0,0):

,5,3,1,;16

1

11

4

2

nmtpa

M

n

mn

m

xy

22

2

222

032,0

0314,0

02877,01

pa

pa

pa

Mnm

t Axymn

B(a,0):

A

xy

B

xymn MMnm

t

222

1

C(0,a):

AxyCxymn MM

nmt

222

1

D(a,a):

A

xy

D

xymn MM

nm

t

222

1

n axele de simetrie:

x=a/2sauy = b/2

Mxy

= 0

ax

y

a

Mx yM

Mmax pa2= 0,047

y

a

ax

xMy

0,032pa

2


27/27

S se determine wmax,Mmax iMxy,max pentru o placptrat din beton armat

simplu rezemat pe contur cu

a= 6 m;E= 3e7 kN/m2; = 0,2; h= 14 cm;p= 20 kN/m2

lund m= n= 7, respectiv 50 i 100.

Sse compare soluiile cu cele furnizate de MEF.

Tem: termen de predareS12 (17.12.2014)

s8 PPD1 Teorie Polin Navier

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