01_02_03_Răsucirea împiedicată

7/30/2019 01_02_03_Rsucirea mpiedicat

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RSUCIREA MPIEDICAT (Torsion with restraint warping)

Un element de construcie este solicitat la rsucire(torsiune) dac n orice seciune transversala sa singurul efort diferit de zero este momentul de rsucire.

OBS. De exemplu, n urma rsucirii unei seciuni dreptunghiulare, seciunea transversal iniialplan trece aproximativ ntr-un paraboloid hiperbolic (vezi figura 1a). La seciunile transversalealctuite din dreptunghiuri, pe lng deplasarea seciunii transversale apare i o strmbare aliniei mediane a seciunilor. De exemplu, linia median a unui profil dublu T, prsete planul

yGz, astfel nct aceasta a devenit o linie strmb n spaiu (vezi figurile 1b, 1c).

z

y

x

Mt

z

y

x

z

y

x

Mt

a) b) c)

Figura 1

Dac fenomenul strmbrii se poate produce liber, rsucirea se numete libersau pur. Dac

strmbrile sunt mpiedicate, total sau parial, rsucirea este calificat ca i mpiedicat.

mpiedicare strmbrii poate fi provocat de rezemarea ncastrat a extremitii barei, de variaiabrusc a seciunii transversale, etc. mpiedicarea strmbrii duce la apariia unor tensiunisuplimentare i complic studiul strii de solicitare.

OBS. n cazul barelor cu seciune robust (seciunea dreptunghiular), mpiedicarea strmbriiseciunii are o implicaie mult mai sczut dect o are mpiedicarea strmbrii liniei mediane labare cu perei subiri.

1.1 Rsucirea mpiedicat a barelor cu perei subiri cu seciune transversal simpluconex (profil deschis)

1.1.1 Fenomenul rsucirii mpiedicate, ipoteze

Se reamintete c n cazul rsucirii barelor cu perei subiri, pe lng strmbarea suprafeeiseciunii transversale, exist i o strmbare a liniei mediane. Se exemplific acest lucruconsidernd o bar cu seciune dreptunghiular ngust (vezi figura 2a), respectiv o bar cu pereisubiri (seciune dublu T) alctuit din dreptunghiuri nguste (vezi figurile 2c, 2d).


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z

y

x

Mtx

z

Mt0

Mt0 Mt0

Mt0

a) b) c) d)

Figura 2:

n urma rsucirii, seciunea transversal se deplaseaz lund forma unui paraboloid hiperbolicnscris ntr-un dreptunghi strmb. Se constat c linia median Gz a seciunii rmne n planuliniial al suprafeei seciunii, ceea ce nseamn c n planul median al dreptunghiului ngustlunecarea specific este nul.

unghiul iniial drept ntre axa Gxi Gznu se modific

00

yxz (1)

Relaia (1) constituie baza teoriei rsucirii mpiedicate.

n figura 2c este reprezentat forma deformat a unui profil dublu T supus la rsucire, prinaplicarea a dou momente de rsucire de capt egale.Pe lng strmbarea propriu-zis a seciuniitransversale se produce i o strmbare a liniei mediane (vezi figura 1).

Figura 2c prezint rsucirea liber, n timp ce figura 2d, pentru bara cu o extremitate ncastrat,prezint o deformaie relativ nou a barei, strmbarea liniei mediane a seciunii transversalevariind pe lungimea consolei de la 0 n dreptul ncastrrii pn la o valoare maxim la captulopus. Rezult c strmbarea liniei mediane dintr-o anumit seciune este mpiedicat, parial sau

total, de tendina de strmbare mai redus a liniei mediane din seciunea vecin ; de aici provinei denumirea de rsucire mpiedicat.

mpiedicarea strmbrii liniei mediane reprezint n fond mpiedicarea deplasrilor elastice alepunctelor liniei mediane dup direcia axei barei, ceea ce determin apariia unor tensiuninormale

.

Deoarece mpiedicarea strmbrii este variabil n lungul barei, rezult c i tensiunile normale

variaz, ceea ce atrage dup sine apariia unor tensiuni tangeniale

, din motiv de echilibru

(vezi forma deformat a consolei din figura 2d).

n urma acestei deformaii, n seciunea transversal apar trei feluri de tensiuni:


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1. Tensiuni normale , provenind din ncovoierea tlpilor. Din figura 2d se observ c

tlpile profilului sunt ncovoiate n planurile lor ca nite console, dar n sensuri contrare.Figura 3a prezint distribuia tensiunilor normale . Se observ c tensiunile normale

sunt n echilibru local. Efectul lor nu se amortizeaz pe lungimi de bar de acelai

ordin de mrime cu dimensiunile seciunii transversale ci se extinde pe toat lungimea

barei. Din acest motiv, pentru cazul rsucirii mpiedicate, principiul lui Saint Venant ipierde valabilitatea i n consecin metode reducerii i pierde valabilitatea.

Pentru a putea pstra ideea reducerii tensiunilor fa de punctele semnificative ale seciuniitransversale, Vlasov a introdus un efort rezultant fictiv numitbimoment, notat cuB.

Pentru o seciune dublu T, bimomentul se poate defini ca i produsul dintre momentul M dinplanul tlpilor i distana dintre planurile mediane ale acestora:

)( thMB [kNm2 , daNcm2] (2)

2. Tensiuni tangeniale , provin ca urmare a variaiei liniei mediane a seciuniitransversale n lungul barei (figura 3b). Rezultanta acestor tensiuni n cele dou tlpi,reduse fa de centrul de tiere(rsucire) C, definesc un moment de rsucire M , numit

moment de ncovoiere rsucire sau moment de rsucire mpiedicat. Indicelesemnaleaz proveniena mrimilor respective din cauza mpiedicrii strmbrii linieimediane i este preluat de la notarea aa numitei coordonate sectoriale.

z

y

xM

M

s ?

z

y

x

M?

t ?

z

y

x

Mti

t i0

a) b) c)Figura 3:

3. Momentul de rsucire M nu este ntregul moment de rsucire tM din seciuneatransversal curent

lt MMM (3)

lM - momentul de rsucire liber. n figura 3c sunt reprezentate tensiunile tangeniale l care

provin din acest efort (vezi rsucire pur liber)

l

M

lM


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Se consider cazul unei seciuni dublu T (figura 4):

Figura 4

Dac n urma rotirii seciunii din rsucirea barei, deplasarea unui punct P al liniei mediane arecomponent dup linia median ce trece prin acel punct atunci n punctul considerat se vaproduce i o strmbare a liniei mediane, deplasarea elastic a punctului va avea i o componentnormal pe suprafaa seciunii transversale; o deplasare component u.

1.1.2Deducerea formulei de calcul a tensiunii normale

Aspectul geometric

Se consider bara cu perei subiri din figura 5.

a) b)Figura 5

Un punct oarecare Pde pe linia median a seciunii poate fi poziionat fie prin coordonatele salecartezieney,z n sistemul de axe de inerie principale, fie prin coordonata curbilinie s n lungulliniei mediane.


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n urma deformrii barei, seciunea transversal se rotete cu un unghi n jurul centrului dersucire C.

OBS: Se consider, ca i convenie, c rotirea este pozitiv dac sensul de rotire este cel orar.

Deplasarea punctului P n sistem cartezian va avea componentele (u, v, w)(x,y,z) sau u,, dup axele x,, , unde primele dou sunt axe intrinseci punctului P (axa este tangent n Pla linia median a seciunii iar axa este normal n P la linia median).

Considernd indeformabilitatea seciunii transversale, din figura 5b, rezult:

rPPPP coscos" (4)

Pentru determinarea componentei u, se utilizeaz relaia (1), adic 00

yxz

ntr-un punct

curent al liniei mediane

0

xs

u

x

(5)

Pentru cazul unei seciuni constante n lungul barei, r nu depinde de x, astfel nct

r

dx

dr

xr

x(6)

Prin integrare relaia (5) devine

1

0

1

0

)()( 11

s

s

s

s

dsrxfdsx

xfu (7)

Integrarea se face n lungul liniei mediane, pornind de la punctul P0 numit punct sectorialprincipal, pn la punctul curent P. Funcia de integrare )(1 xf este nul deoarece ar corespundeunei translaii rigide a seciunii transversale, incompatibil cu deformaia barei din rsucire.

Completnd figura 5 cu figura 6 se observ c dsrd , adic dublul ariei triunghiuluielementar msurat. Mrimea notat cu reprezint coordonata sectorial a punctului P.Coordonata sectorial este pozitiv dac poziia razei vectoare a punctului respectiv se obineprin rotirea razei vectoare origine CP0 n sens orar.

Alungirea specific

xu

x- aspectul geometric (8)

Limitnd deformaiile barei la domeniul de comportare elastic, pe baza legii lui Hooke EE xx - aspectul fizic pentru domeniul elastic de comportare (9)

Nu se cunosc: funcia rsucirii , poziiile punctelor C i P0 pentru definirea coordonateisectoriale .


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Figura 6

Aspectul static din figura 5 se observ c singurul efort diferit de zero este momentul de rsucireMt.

Sinteza aspectelor, const n identificarea eforturilor calculate static i pe cale de rezisten.

a) A A

x SEdAEdAN 00 (10)

A

dASE 00 0 (11)

0S [L4

, de exemplu cm4

] - momentul static sectorial al seciuniib)

A A

zy SEdAzEdAzM 0

A

zdAzS 0 (12)

c) 0 0z yA

M S y dA (13)

Sz, Sy [L5, de exemplu cm5]momente statice sectoriale liniare

Condiiile (11), (12), (13) permit definirea complet a coordonatei sectoriale : ultimele dou

precizeaz coordonatele carteziene ale centrului de rsucireC

, iar prima determin coordonatacurbilinie s0 a punctului sectorial principal P0.

Deoarece, bimomentulB nu poate fi definit static, Vlasov l-a exprimat sub forma:

A

dAB (14)

sau A

IEdAEBE 2

unde A

dAI2

I [L6, de exemplu cm6]moment de inerie sectorial


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I

B

I

BE

OBS. BimomentulBnu se poate calcula pe cale static.

Exemplu de determinare a bimomentului pentru o seciune dublu T

Figura 7

A A A

dArydArdMthMB )(

1.1.3Deducerea formulei de calcul a tensiunilor tangeniale

Se izoleaz un element diferenial prin dou seciuni traversale infinit vecine i un planlongitudinal care la nivelul s de secionare este normal pe linia median a seciunii transversale(figura 8).

Tensiunile normale de pe suprafaaAs definesc rezultanta:

As

sdAI (15)

Datorit variaiei tensiunilor n lungul barei, n seciunea x+dx, atunci i rezultanta Is va avea o

cretere cu cantitatea diferenial dIs.

Pe faeta longitudinal, tensiunile tangeniale se nsumeaz n rezultanta:

dxtdL ss (16)

Se consider c este uniform distribuit pe limea ts.

Din condiia de echilibru pe axaxrezult:

ssssss dLdIdLIdII 0 (17)

Din relaia

As

sI

SBdA

I

BI

I

B

(18)


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unde: As

dAS - momentul static sectorial al poriunii de arieAs.

Figura 8

Prin difereniere, innd cont c dBI

SdIconst

I

Ss

.

Ecuaia (17) devine:

It

SM

It

Sdx

dB

dxtdBI

S

ss

s

(19)

undedx

dBM - momentul de ncovoierersucire (20)

OBS. Relaia (19) are structura matematic a formulei lui Juravski.

1.1.4 Ecuaia diferenial a rsucirii mpiedicate, integarea ei

Din ecuaia lt MMM

- momentul de rsucire liber,Ml

l t tM G I G I (21)

dx

d - rsucirea specific (Rotirea relativ n jurul axei barei a dou seciuni situate la

distan unitar, se numete rsucire specific ()).

- momentul de ncovoierersucire,M IE

dx

dBM (22)


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B E I (23)

t tM E I G I (24)

Se deriveaz ecuaia (24) n raport cu abscisa barei:

tIV

t

t IGIExm

dx

dM)( (25)

Se mparte relaia (25) cu EIi se noteaz cu:

IE

IGk t

(26)

Astfel c relaia (25) devine:

IE

xmk tIV

)(2 (27)

Ecuaia (27) reprezint ecuaia diferenial a rsucirii mpiedicate.

Soluia acestei ecuaii este:

pg (28)

unde: g soluia general a ecuaiei omogene;

p soluiaparticular.

Soluia devine:

0 0 0 0

1' t p

t t

shkx chkx kx shkxB M

k GI kGI

(29)

unde:

0 0, ' - rotirea seciunii, respectiv rsucirea specific n dreptul originii barei (z = 0);

B0bimomentul n origine;Mt0momentul de rsucire concentrat n aceast seciune.

Prin derivarea soluiei, aceasta devine:'

00

'

0

1p

t

t

tGI

chkxM

GI

kshkxBchkx

(30)

Prin derivarea relaiei (30) i multiplicnd cu (-EI) se obine relaia bimomentului:''

00

'

0 ptt EI

k

shkxMchkxBshkx

k

GIB (31)

Soluia particular p se determin astfel:

se caut o funcie x care s satisfac condiiile0)0(")0(')0( ; 1)0(

o astfel de funcie este:

34k

kxshkx

(32)

se face o schimbare de variabil x x-t


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se alctuiete soluia particular, folosind funcia4

i membrul drept al ecuaieidifereniale

x

t

t

x

t

pdttmtxktxshk

kGIdt

EI

tmtx

00

4 )()()(1)(

)(

(33)

Se observ c soluia particular depinde de modul de ncrcare a barei.

mt(t) = mt

Mtc

a) b)

Figura 9: Dou tipuri de ncrcare ntlnite n practic

Pentru ncrcarea cu moment de rsucire uniform distribuit (figura 9a)axo 0p

bxa

2)(

22

2

axk

laxchkGIk

m

t

t

p (34)

lxb

2

)()(

2

)()(

2222bxk

bxchkaxk

axchkkGI

m

t

t

p

Pentru ncrcarea cu moment concentrat (figura 9b)cx 0 0p

lxc )( cxkcxshkkGI

M

t

tc

p (35)

Pentru ca funciile , i B s devin operante trebuie s cunoatem valorile parametrilor

iniiali 0'

00 ,, B i 0tM , parametrii care se determin pe baza condiiilor de margine (rezemare).

Astfel avem:a) capt simplu rezemat- din punct de vedere al rsucirii este acel reazem care mpiedic rotirea seciunii, dar

permite stmbarea liniei mediane. Astfel, prin strmbarea liniei mediane (liber s seproduc) tensiunile normale 00 B . Condiiile pentru rezemarea simpl se scriu

astfel:

0

0

0

0

B


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b) capt ncastrat- aceasta va mpiedica att rotirea seciunii ct i strmbarea liniei mediane care este obligat

s se menin n planul seciunii ncastrate. Astfel condiiile sunt:

0

0'

0

0

u

Pentru orice valoare a coordonatei sectoriale (pentru orice punct la secinii mediane), rezultc: 0'0 .

c) capt liber (nerezemat)Condiiile se pun asupra eforturilor:- dac marginea este nencrcatB0 =Mt0 = 0- n caz contrar, eforturile primesc chiar valorile ncrcrilor direct aplicate.OBS: n cazul unor rezemri intermediare pe lng condiiile la limit puse n dreptul capetelorbarei, sunt necesare i condiiile de continuitate a deformaiilor, respectiv de echilibru n dreptulreazemului intermediar.

1.1.5 Detreminarea caracteristicilor geometrice sectoriale

A. Determinarea poziiei centrului de rsucireCentrul de rsucire se determin n dou etape:1) Se alege un centru de rsucire arbitrar C1 i un punct sectorial principal, de asemeneaarbitrar

'

0P .2) Se face translaia de la centrul arbitrar C1 la adevratul centru de rsucire C, folosindrelaiile (12, 13).

n figura 10 s-a notat cu raza vectoare CP, cu rperpendiculara dusa din Cpe tangenta n P lalinia median, cu unghiul dintre raza vectoare CQi axa de inerie Gy. Cu aceste notaii, sepoate scrie difereniala coordonatei sectoriale definit prin utilizarea centrului de rsucire Ci apunctului sectorial principal arbitrar Po.


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Figura 10

Axeleyizsunt axe de inerie principale Sy = Sz =Iyz=0, astfel c se permite explicitareaabsciseloryCizCale punctului C. Rezult formulele de calcul ale coordonatelor centrului dersucire:

'

1

1

'

1

1

' 0

' 0

Ay c c

yA

Az c c

zA

z dA

S y dA y yI

y dA

S z dA z zI

(36, 37)

unde,'1 - coordonata sectorial a punctului curent, definit cu ajutorul centrului de rsucire arbitrarC1

i cu un punct sectorial principal arbitrar '0

P

Pentru seciuni la care linia median este alctuit din linii drepte, integralele din relaiile (36,37)se pot rezolva cu regula de integrare Vereciaghin. Integralele de suprafa se pot rescrie subforma unor integrale liniare dA=ts.ds:

'

1

1

'

1

1

i

i

i

s

c c

y

i

s

c c

z

t z ds

y yI

t y ds

z zI

(38, 39)

unde suma de la numrtor se extinde pe numrul de lungimi de linie median si cu grosimeaconstant tirespectiv.

B. Determinarea poziiei punctului sectorial principalCu raza vectoare origine '0CP , coordonata sectorial a punctului P este dublul ariei CPP

'

0 .

Coordonata sectorial corect a aceluiai punct Ptrebuie ns definit plecnd de la raza origineCP0, coordonata fiind dublul ariei CP0P. Rezult c: d ,

unde: - coordonata sectorial definit de raza '0CP ;


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d - coordonata sectorial de diferen, adic dublul ariei 0

'

0 PCP .

?2

? d2

Figura 11

Folosind condiia de mai sus se obine:

A A A

d dAdAdAS 00

A

dAA

d

(40)

C. Calculul momentului de inerie sectorialCunoscnd poziiile punctelor C i P0, coordonata sectorial a punctului P este determinat.Reprezentarea variaiei coordonatelor sub forma unei diagrame, permite n cazul frecvent alseciunilor cu linia median alctuit din linii drepte, calculul momentului de inerie sectorialprin

procedeul de integrare Vereciaghin a diagramelor:

Si

i

A A

sdstdstdAI

2 (41)

OBS. Toi termenii sunt pozitivi; I [L6]moment de inerie sectorial.


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Calculul practic al momentul critic elastic

Pentru cazul uzual al grinzilor cu seciune constant, dublu -simetrice sau mono-simetrice, nraport cu axa minim de inerie, momentul critic elastic poate fi determinat cu expresia (42).Aceast expresie este aplicabil elementelor structurale ncovoiate dup axa maxim de inerie,

pentru diverse rezemri i tipuri de ncrcare.

0.522 2

21 2 3 2 32 2

( )( ) ( )

( )

wz z z Tcr g j g j

w zz z

IEI k k L GIM C C z C z C z C z

k Ik L EI

(42)

Coeficientul kzse refer la posibilitatea de rotire a seciunii transversale pe reazeme, dup axaminim de inerie a seciunii, iar coeficientul kw se refer la posibilitatea deplanrii seciuniitransversale pe reazeme. Aceti coeficieni variaz ntre 0.5 (fixare perfect la ambele capete),0.7 (fixare perfect la un capt i celalalt capt liber) i 1.0 (liber la ambele capete). Dac nu s-auluat msuri speciale pentru fixarea deplanrii seciunii transversale n dreptul reazemelor,

coeficientul kw poate fi considerat, n mod conservativ, egal cu unitatea. Dealtfel, avnd nvedere c n multe situaii practice fixarea att din punct de vedere al ncovoierii dup axaminim de inerie ct i din punct de vedere al deplanrii este doar parial, ambii coeficieni potfi considerai n mod conservativ egali cu unitatea. Cu toate acestea, exist detalii structurale dembinare sau de rezemare a grinzilor pentru care se poate considera o fixare perfect.Proiectantul trebuie s aib n vedere relaia ntre modul de alctuire al detaliilor structurale ialegerea coeficienilor lungimilor de flambaj pentru calculul momentului critic elastic pentruflambaj prin ncovoiere rsucire. n Figura 12 sunt prezentate cteva exemple cu detaliistructurale (mbinare rigid i articulat rigla-stlp, mbinare articulat grind secundar grind

principal, reazem articulat grind) pentru care sunt precizate condiiile de fixare.

- ncastrat pentru ncovoiere dup axa y - Articulaie pentru ncovoiere dup axa y- Rotire dup axa z mpiedicat - Rotire dup axa z liber- Deplanarea mpiedicat - Deplanare libera) mbinare rigid grind stlp b) mbinare articulat grind stlp

- Articulaie pentru ncovoiere dup axa y - Articulaie pentru ncovoiere dup axa y- Rotire dup axa z liber - Rotire dup axa z parial mpiedicat- Deplanare liber - Deplanare liberc) mbinare grind secundar-principal d) Reazem articulat-seciune de capt liber


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- Articulaie pentru ncovoiere dup axa y- Rotire dup axa z mpiedicat- Deplanare mpiedicate) Reazem articulatseciune de capt rigidizat

Fig. 12: Detalii de mbinri


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Tabelul I.1: Caracteristicile sectoriale pentru diferite seciuni uzuale

Nr.crt.

SeciuneaAria, centru de greutate,

momente de inerieCentru de rsucire

Momentul de ineriesectorial

2 2y zr dAy I

y

0 1 2 3 4 5

1

t tz

b bC

G y.zC

btA 2 24

bz e

c

0

I

2

2

bry

innd seama de

grosimea pereilor3 3 3

18 144

b t AI

2

t

t

z

.

C

G y.zC

.yC

.

b1

b2

21 bbtA 32313

36bb

tI

Seciunea contraindicatpentru flambaj prinncovoiere-rsucire

3t

z

b

C

Gy

.e.

zCA

aa .ta Aa,Ia

2 aA bt at ;

aa atA

2

2 3 26

ac

z

t baz b aI

2

2

2 a

c z

I b I

z e I

4

2

10

222

2

2

4

2 3

3 2 10

y

y

a

y

tbr

I

b e t a b a

Ib e

b a

ab a e

31

3a aI a t


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4 t

hG=C y

b

z

22

1

1

2A t b h 0y zC C

3 2

6

b h tI

sau2

14

hI I

0y zr r

5

z

G y

h

C

.tb

thAh,Ih .zce

b

b bA t b ; h hA t h

2h bA A A ;bAe b

A

0cy

bc

z

Iz e b

I

22 2

3

b h b

z

I I IbI

I

2

44

1

2

2

y y hy

b

b

r e A e I I

e b I

te b e

2

2b b

hI A

;

3

12

hh

t hI

2z h bI I I ;

3 22

3y bI b t e A

6

z

A y

C

. th

tiAb,Ib.zc

e

h

.

Ah,Ih

Aa,Ia

Gb.t2

b

.b2

c

h hA t h ; 1 1 1A t b

2 2 2A t b

1 22 2hA A A A

22hA Ae hA

2

2h h

bI A

;3

1 1

1 12

t bI

232 2

2 212 2

t b bI A c

322 2

2 212

h

t bI A c

2 12 2z hI I I I

3 2 22

22

3y hI h t A h e A

0cy ;2

21 24

hc

z z

Ib Az e h

I I

22

22

2

2 2 22

22 2

4

1 24

2

4

y

hz

h

z

h

z

bI I e A

b Ah I

I

Ibch A b heA

I

Ih

I

21 1

44

2

2

2

1

2

2

2

y

y

h

h

r e A e I I

e h I

te h e

h e I

A h e


18/28

7

z

A y

b

C

.th

tbAb,Ib.zc

eh

.

Ah,Ih

Aa,Ia.ta

G

a

a aA t a ; b bA t b

h hA t h ;2 2h b aA A A A

2h aA Ae hA

0cy ; cz e 2

12

4ah

z

eAbhI

I

22

2

2 2

2

4

14

2 2

y

z

ah a

ah

z

bI I e A

b A

I

h I bfh A

Ib e Ah

I

2

2h h

bI A

;

3

12

bb

t bI ;

3

aah

tI a

23

12 2

aa a

t a b aI A

2 2z h b aI I I I

3 2 222

3y h aI h t A h e A

8

z

A y

b

C

.th

tbAb,Ib.zc

eh

.

Ah,Ih

Aa,Ia

taG

a

.

La calcularea valoriiaI ,

semnele se iau astfel:semnul (+) pentru cazul

7semnul ( ) pentru cazul

8

/ 2f a / 2f a

9

z

yh

t G=C

b

.

.ti

0cy ; 0cz 2

4z

hI I 0y zr r


19/28

10

z

y

G=C h

t

. bt

Aa,Ia

Ah,Ih

.ba

.

.tt

ta

e

Ab,Ib

h tA h t t ;

b t a t A b t t ; a a aA b t ;2 4h b aA A A A

3

12

th

t h tI

;

3

12

b t ab

t b tI

;

3

12a aa

t bI ;

2

2 42

tz b a

bI I A

2

4

2 42

y h a

b a

I I I

hA A

0cy ; 0cz

22

242 2

z a

bhI I I

0y zr r

11

z

y

h

ti

G

.t

.

.

ts

C

A

. ezc

.bs

bi

Ah,Ih

Ai,Ii

As,Is

1

2h s iA h t t t

;

sss tbA ; i i iA b t

h s iA A A A ;2

2

h iA Ae hA

;

3

1

2

12

s i

h

t h t t

I

;

3

12

s ss

t bI ;

3

12

i ii

t bI ;

z s iI I I 2 24y h iI I A h e A

0cy

1

c s iz

z eI h e II

2i s

z

I II hI

3

3 4

4

1

4

y c z sz

i

r z I A eI

tA h e e

h e


20/28

12

z

y

.hiG

.ts C

zc=e

.bs

ti. h

.

s s sA b t : i i iA h t

s iA A A ;2

iAe hA

3

12

s sz

t bI ;

32

3

i iy

t hI e A

0cy ; cz e

0I Cu luare n

considerare a grosimiipereilor

3 3 3 3

144 36

s s i it b t hI

3

344

1

4

y z sy

i

r I e A eI

te h e

13

z

yG=C. t2

.t1

h

b

2 21 2

2 1

2r

b h t t I

bt ht

0c cy z

22 22 1

2

2 1

1 2

24

bt ht b hI

bt ht

bt ht

0y zr r


21/28

3.8.6 Torsional moment

In light gauge steel sections, the role played by torsion is enhanced in comparison tostockier hot-rolled sections. There are three reasons why this is so (Davies, 1991):1. Most cold-formed members have an open cross-section and often have a shear centre

which lies outside the section. It follows that application of the load through the shear

centre is difficult to achieve and cold-formed beams are, in fact, subjected to an appliedtorsion so that twisting of the section occurs if it is not restrained against torsion;

2. For open cross-sections, the torsional constantIt, which is directly related to the resistanceto twist, is proportional to the material thickness raised to the third power. It follows thatthin walled cold-formed sections have less resistance to torsion than hot-rolled ones;

3. When torsion is restrained, for instance at the supports, longitudinal stresses appears whichmay be of the same order of magnitude as the bending stresses and should be added tothem. A section that is designed to be fully stressed in bending may therefore beoverstressed when restrained torsion is taken into account.Hence the load-bearing resistance is reduced substantially by torsion so that torsional

moment should be avoided in construction. However, in practice, light gauge steel beams areoften loaded through the members that they support and these, in turn, provide torsional restraint.In the most frequent applications (e.g. purlins, floor joists, wall studs) the loading and restraintare both continuous and to some extent self-equilibrating so that the tendency to twist is greatlyreduced.

In design the superposition of stresses due to axial force, bending moments and torsionalmoments must remain below the limit of yield stress. Additionally, the superposition of shearstresses has to be proved.

The elementary theory of torsion is usually known as St. Venant's theory. For a memberunder uniform torsion, the angle of rotation per unit length is related to the torsional momentthrough the following equation:

t

t

T d

GI dx

(3.93)

whereTt= is the torsional moment;

It= is the torsion constant;G = is the shear modulus;= is the twist of the section over a length;

x = is a variable with the direction of the longitudinal axis of the member.The torsional constant It is given with sufficient accuracy for all engineering purposes for

an open cross-section of uniform thickness tby

3

3t

l tI

(3.94)

where l is the length of the middle line of the cross-section.For a closed cross-section, the torsional constant tI is:

24t

A tI

l (3.95)

whereA is thearea enclosed by the mid-thickness line.Equation (3.93) is concerned solely with shear stress and this defines the restriction on the

St. Venant theory for it is based on the fundamental assumption thatplane sections remain plane.This assumption holds for circular sections only but, unfortunately, for both open and non-circular closed sections it is not acceptable. The cross-sections of light gauge steel members in


22/28

torsion do not generally remain plane but undergo warping. It is convenient to illustrate the basicphenomenon with reference to a doubly symmetrical I-section as shown in Figure 3.49.However, warping is of more general occurrence in sections of all types.

Figure 3.49(a) shows a member subjected to uniform torsion with no restraint to warping.The member twists, as shown in plan in Figure 3.49(b), and only the web remains plane whilethe flanges rotate bodily. The section as a whole thus departs from its original plane cross-

section and this type of deformation is known as warping. If no restraint is provided to resist thisdeformation then no additional longitudinal stresses are developed.If this warping is restrained, for instance the case of a cantilever (see Figure 3.49(c)), the

flanges are forced to bend in the horizontal direction. This bending of the flanges is clockwisefor one flange and anti-clockwise for the other so that the effect is of two equal and oppositemoments. This type of behaviour was the subject of a classical investigation by Vlaslov whotermed this force system induced in the flanges by warping restraint a bimoment.

(a) (b) (c)Fig. 3.49: Unrestrained and restrained warping: (a) Member subject to uniform unrestrained torque, (b) Plan view of

unrestrained torque, (c) Plan view of restrained torque showing componentsMof the bimoment (Davies, 1991)

The bimomentB has units of force distance2 and it can be considered as a combination ofthe momentsMand the depth h of the web as shown in Figure 3.49(c). Thus

B M h (3.96)

The presence of a bimoment causes longitudinal and shear stresses in the section. Thelongitudinal stresses act in the same sense as the ordinary flexural stresses and may be of thesame order of magnitude. It follows that consideration of the combined longitudinal stresses dueto both bending and warping is often necessary.

The longitudinal stress associated with a bimomentB is given by:

w

B

I

(3.97)

where Iw is a property of the cross-section termed the warping constant and is the sectorialcoordinate which varies around the section.

Values ofIw and for some common sections, together with It and the position of theshear centre are shown in Table 3.9.

For sections that are not included in Table 3.9, a general method to calculate the torsionalproperties is presented in Annex A.

Therefore, eqn. (3.96) makes possible the determination of the longitudinal stresses due torestrained torsion provided that the bimoment, warping constant and sectorial coordinates areknown. The latter two quantities are properties of the cross-section but the bimoment is afunction also of the loading and support conditions and must be evaluated separately for eachgiven situation.

Uniform torsion induces distortion that is caused by the rotation of the cross-sectionsaround the longitudinal axis. As a consequence, shear stresses appear which balance the applied


23/28

torsional moment T; under these circumstances, the resistance to the torsional moment Texclusively results from St. Venants torsion, Tt. Although longitudinal warping displacementsmay exist, they do not introduce stresses.

In non-uniform torsion, besides the St. Venant shear stresses, longitudinal strains also exist(because warping varies along the member). These longitudinal strains generate self-equilibrating normal stresses at the cross-sectional level that, depending on the level of

restriction to warping, vary along the member. The existence of varying normal stresses implies(by equilibrium in the longitudinal direction) the existence of additional shear stresses that alsoresist to torsional moments, leading to:

t wT T T (3.98)

Table 3.9: Torsional properties of some common cross-sections

WarpingconstantIw

TorsionconstantIt

e for shearcentre

Sectorial coordinate values

bf

bwO

tw

tf

2 32

or4 24

w f fz wb b tI b

3 3

23

f f w wb t b t

or if w ft t

3

23

f w

tb b

0

(bfbw)/4

(bfbw)/4

bf

bwO

e

t

2 3

32

6121

f

w f w

f

w

b

b b t b

b

b

3

23 f w

tb b

23

6

f

f w

b

b b

[(bf -e)/2]bw

(ebw)/2

bw

O

e

a

a

t

bf

2

3 2(4 3

6

f

w

b ta b a

3 2

2

6 )w f w

y

b a b b

I e

3

23

2

f

w

tb

a b

2

2

2

1 2

2 4 3

f w

y

f

w

b bat

I

b a

a b

[(bf -e)/2]bw

ebw/2

-{[(bf -e)/2]bw+

+[a(bf+e)]}

[(bf -e)/2]bw

bw

bf

a

O

t

2

2 2

2 2

2

12 2 2

[ ( 2

4 6 )

4 (3 3

4 2 )]

f

f w

w f f w

f w

f w w

f w

b tb b a

b b b b

b a b a

a b b b

b a b a a

3

2 23

f

w

tb a

b

0

-bwg/2

g

(bf -g)bw/2

(bf -g)bw/2-abf

where

g=bf /(bw+2bf )

bf

O

t

Obw

3 2 2

12 2

f w f w

f w

b b t b b

b b

3

23

f w

tb b 0

-bwg/2

g

+(bf -g)bw/2

where

g=bf /(bw+2bf )


24/28

Obf

bw

t

3

3 3

36f w

tb b

3

3w f

tb b 0

Values of for this sectionare secondary effects only,

for the upper and lowersurfaces

The applied torsional moment T is thus balanced by two terms, one due to the torsionalrotation of the cross-section, Tt, and the other caused by the restraint to warping, designated bywarping torsion, Tw.

The equations relevant to this problem will first be derived for a thin-walled I-section andthen generalised. From eqn. (3.93) results:

t t

dT GI

dx

(3.93)

The second part of the torque, Tw, is found by considering the bending of the flanges due towarping. Using the symmetry of the section to note that each cross-section will rotate about the

x-axis as shown in Figure 3.50, allows the lateral deflection of the flanges v to be written as:

2

hv

(3.99)

The lateral bending moment in each flangeMfwill be:

2

2f f

d vM EI

dx (3.100)

whereIfis the second moment of area of the flange about thez-axis. Substituting for v gives2

22f f h dM EI

dx (3.101)

v

h/2

y

Vf

V

f

Fig. 3.50: Warping torsion of an I-section

The shear forces across the width of each flange Vfwill be given by:

3

32

f

f f

dM h dV EI

dx dx

(3.102)

The couple produced by these two shear forces, provides the second part of the resistanceto twist Tw. Thus


25/28

2 3 3

3 32w f f w

h d dT V h EI EI

dz dz

(3.103)

where, for a thin-walled I-section 2f yI I and2 2

2 4

f y

w

I h I hI is the warping constant and

wEI is the warping stiffness of the section.Combining eqns. (3.98), (3.93) and (3.103) results

3

3t w t w

d dT T T GI EI

dx dz

(3.104)

This is the general equation for the torsion of a non-circular section and the evaluation ofthe longitudinal stresses and shear stresses due to an applied torsion requires its solution withappropriate boundary conditions. Differentiating gives the alternative form.

4 2

4 2w t

dT d d EI GI m

dx dx dx

(3.105)

or4 2

2

4 2w

d d mk

EIdx dx

(3.106)

where tw

GIk

EI and m is the intensity of a uniformly distributed torsion (m = 0 for a

concentrated torsional moment).It may be noted that it also follows from eqn. (3.101) that bimoment is related to twist

according to:

2

2w

dB EI

dx

(3.107)

In order to calculate values of bimoment and then to evaluate the stresses arising fromrestrained warping it is necessary to obtain solutions of the general equation of torsion.

The general solution of eqn. (3.106) is, therefore,

1 2 3 4 0cosh sinhC kx C kx C x C (3.108)

where

C1 to C4 are constants;0 is the particular solution = 0 for m = 0, or

2

2 t

mx

GI for uniform m.

The values of the constants C1 to C4 are determined by substituting known boundaryconditions into eqn. (3.108).

The two most common idealised support conditions are:(1) Fixed endone which is built in and can neither twist nor warp, i.e.

0 , 0d

dx

(3.109)

(2) Simply supported endone which cannot twist but is free to warp and is therefore free oflongitudinal stresses due to torsion


26/28

0 ,2

20

d

dx

(i.e.B = 0) (3.110)

Although it is possible to obtain these idealised conditions in practice, the supportconditions that are usually provided fall somewhere between the two so that accurate boundaryconditions are difficult to specify. The values of bimoment for common boundary conditions are

presented in Table 3.10.

Table 3.10: Bimoment variation for common boundary conditionsLoad condition Bimoment equation Maximum values

M

sh ( )

ch

k l xB Ml

kl kl

0x B Mlb

m

2

[ shch

ch ch ]

mB kl k l x

k kl

kl kx

0x 2B ml c

m

2

ch2

1

ch

2

lk x

mB

klk

2

lx 2B ml p

M

sh

2ch

2

M kxB

klk

2

lx

2

MlB f

m

2

ch2

1

2 sh

2

lkl k x

mB

klk

0x

x l

2

lx

2B ml g

2B ml j

M

ch ch2

2sh

2

lkx k x

MB

klk

0x

2

lx

x l 2

MlB n


27/28

M

2

2 2

[1 ch

1 sh ch2 sh

ch

mB kx

k

k lkl kl kl

kxkl kl sh kl

x l 2

mlB w

m

1

1

sh sh

ch sh sh2 2 2

MB

k kl kl kl

kl kl klkl kx

2

sh

ch sh

ch sh sh2 2 2 2

M kxB

k kl kl kl

kl kl kl lkl k x

2

lx

x l

2

MlB v

2

MlB u

kl

klb

th ;

kllk

klklklc

ch

1chsh

22

;

2ch

12

ch

22 kllk

kl

p

;

klkl

klf

sh

1ch

kllk

klklkl

gsh

sh)1ch(2

22

;

kllk

klklkl

jsh

2chsh

22

;

)1ch(

2sh2sh

klkl

klkl

n

)shch(

2sh1chsh

klklklkl

klklklklkl

v

;

)shch(

2ch2sh

klklklkl

klklklw

;

klklkl

klkl

ushch

2sh2sh

In conclusion, where loads are applied eccentric to the shear centre of the cross-section, theeffects of torsion shall be taken into account. As far as practicable, torsional moments are bestavoided or reduced by restraints, because they substantially reduce the load bearing capacity,especially in case of open sections.

In determining the effects of the torsional moment, the centroidal axis, shear centre andimposed rotation centre, should be taken as those of the gross cross-section.

The direct stresses due to the axial force NSd and the bending moments My,Sd and Mz,Sdshould be based on the respective effective cross-sections used in 3.8.2 to 3.8.4. The shearstresses due to transverse shear forces, the shear stress due to uniform (St. Venant) torsion andthe direct stresses and shear stresses due to warping, should all be based on the properties of thegross cross-section.

In cross-sections subject to torsion, the following conditions should be satisfied:

, 0/tot Ed y M f (3.111)

,0

/ 3ytot Ed

M

f

(3.112)

2 2, ,

0

3 1.1y

tot Ed tot Ed M

f

(3.113)

where:tot,Ed is the total direct stress, calculated on the relevant effective cross-section;

tot,Ed is the total shear stress, calculated on the gross cross-section.

The total direct stress tot,Edand the total shear stress tot,Edshould by obtained from:


28/28

tot,Ed= N,Ed+ My,Ed+ Mz,Ed+ w,Ed (3.114a)

tot,Ed= Vy,Ed+ Vz,Ed+ t,Ed+ w,Ed (3.114b)where:My,Ed is the direct stress due to the bending momentMy,Sd;Mz,Ed is the direct stress due to the bending momentMz,Sd;

N,Ed is the direct stress due to the axial forceNSd;w,Ed is the direct stress due to warping;Vy,Ed is the shear stress due to the transverse shear force Vy,Sd;Vz,Ed is the shear stress due to the transverse shear force Vz,Sd;t,Ed is the shear stress due to uniform (St. Venant) torsion;w,Ed is the shear stress due to warping.

01_02_03_Răsucirea împiedicată

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