Proiect Constructii Metalice g.

Universitatea Tehnica “Gheorghe Asachi” Iasi Facultatea de Constructii

Proiect la constructii metalice

Student: UNGUREANU DANUT Grupa: 3502

Date personale:

- localitatea de amplasare - Radauti- LH=24 m- LT=6 m- H=8 m

Tip pod PC 32 – gr. Functionare III - incalzita STAS 800/82 - caracteristici pod rulant

A=5600 mm B=7100 mm h1=2455 mm l1=375 mm C1=750 mm b=80 mm masa neta pod Gp=32000 Kg=32000 daN QC=25 KN Lp=18 m P1=22 t.f. P2=8 t.f. P3=27 t.f. P4=10,5 t.f.

STABILIREA DIMENSIUNILOR ELEMENTELORSECTIUNII TRANSVERSALE:

stabilirea dimensiunilor grinzii cu zabrelehc=(1/7÷1/10)L=(1/7÷1/10)*30000=(4286÷3000)mmp=tg α= 590/14600=0,04 α=2,31p%=4% → hm=2660 mm hC=3150 mm α=2,31° a1=a2=a3=a4=2950 mm a5=3000 mm

stabilirea dimensiunilor luminatoruluiLl=6000 mm αl=30°÷45°tg αl=hl/(Ll/2)=hl=tg αl=Ll/2=tg (30°÷45°)6000/2=(1730÷3000)mm

stabilirea dimensiunilor stalpului hS=hgr+b+hl+ht+(300÷400) mm +hm

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hgr=(1/8÷1/10)LT=(1/8÷1/10)*10500=(1312,5÷1050)mm hgr=1200 mm b=90 mm hl=2575 mm ht=(200÷300) mm STAS 800/82 hm=2660 mm hS=1200+90+2575+300-400=3765 mm+2660=6425 mm -rot 50 mm → hS=6500 mm

* hi=H-(hgr+b)+hi=12000-(1190+90)+1400=12210 Hi=(400÷1500) mm rot 50 mm → hi=12200 mm

Podul rulant PC 32 GF IIIA=6300 mmB=7900 mmLp=28 mQC=25 KNQp=45000 daNHS=hS+hi=12200+6500=18700 mm

- apasarea pe rotile caruciorului:P1=22,5 t.f.P2=8 t.f.P3=27 t.f.P4=10,5 t.f.

bS=(1/10÷1/20)hS=(650÷325) mmbS=(400;450;500)Lp+2C-L=0 → C=(L-Lp)/2=(30-28)/2=1000 mmL=30000 mm=30 mLp=28000 mm=28 mC≥bS/2+400+p+l1

bS≤2(C-400-p-l1)=2(1000-400-40-375)=370 mmC1=1000 mm=bS/2+400+p+C1=370/2+400+40+375=1000 mmbi≥(1/20)HS=(1/20)·18700=935 mmbi≥bS/2+C=370/2+1000=1185 mm →bi=1200 mm

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CALCULUL SI ALCATUIREA PANEI DE ACOPERIS

LT=10,5 mm →pane cu 2 tiranti

Stabilirea incarcarilor

-incarcari pe m2 de suprafata aferenta paneiA) PERMANENTE1) greutate invelitoareginv

n=(13 ÷15)daN/m2 – pt invelitori izolate termic

ginvn=13daN/m2 ninv=1,2

ginvc=ninv* ginv

n=1,2*13=15,6 daN/m2

2) greutatea propriei pane:gpn

n=(15÷30)daN/m2

gpnn =(1,10÷1,15)(15÷30)=(16,5÷34,5)daN/m2

gpnn=25 daN/m2 np=1,1

gpnc=hp* gpn

n=1,1*25=27,5 daN/m2

-functie de LT,a → g=250÷300 daN/m2

3) greutate contra vantuiri gcv

n=(2 ÷4) daN/m2 gcvn=4 daN/m2 ncv =1,1

gcvc= hcv* gcv

n=1,1*4=4,4 daN/m2

B)CVASIPERMANENTE 1)Instalatii electrice, conducte, cabluri: gc

n=(1 ÷3) daN/m2 gcn=3 daN/m2 nc=1,2

gcc=nc* gc

n=1,2*3=3,6 daN/m2

2) Incarcari date de praf industrialgpi

n=(1 ÷25) daN/m2 gpin=5 daN/m2 npi=1,4

gpic= npi* gpi

n=1,4*5=7 daN/m2

Incarcari totale

gtn=ginv

n+gpnn+gcv

n+gcn+ gpi

n=13+25+4+3+5=50 daN/m2

gtc=ginv

c+gpnc+gc

c+ gpic+gcv

c=15.6+27.5+4.4+3.6+7=58,1 daN/m2

C) INCARCARI VARIABILE1) Incarcarea din zapada (STAS 10101/21-92) Pz

n=Ce*Czi*gz(KN/m2)

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gz-greutatea de referinta a stratului de zapada(KN/m2) Ce-coeficient prin care se stine seama de conditiile de expunere a constructiilor Czi-coeficient prin care se tine seama de aglomerare de zapada pe suprafata constructiei expusa zapazii Vaslui gz=1,5 KN/m2- perioada de revenire 10 aniCe=0,8-pt conditii normale de expunere si acoperisuri cu profil plat sau putin inclinat Czi=1,0 0≤α≤30° -Czi=1,0 α=2,31° pz

n=Ce*Czi*gz=0.8*1.0*1.5=1.2KN/m2=120daN/m2

clasa de importanta a cladirii IIIzona Vaslui → γa=2,2 γc=1,4perioada de revenire 10 aniγFz=γa-0,4gp/(Ce*gz) ≥0,3γa

γFz=2,2-0,4*50/(0,8*150)=2,033>0,3γa>0,3*2,2=0,66pz

c=γFz*pzn=2,033*120=243,96daN/m2

γ0z=γc-0,2gp/(Ce*gz) ≥0,3*γc*γ0z=γc-0,2*gzn/(C*gz) ≥0,3 γc

γ0c=1,4-0,2*50/(0,8*150)=1,317>0,3γc=0,3 *1,4=0,42pz

a(n)= γ0z*pzn=1,317*120=158,04daN/m2

2)Incarcaridin actiunea vantului (STAS 10101/20 -90) pn

n= β *Cni*ch(z)*gv

β- coeficient de rafala Cni-coeficient aerodinamic pe suprafata i ch(z)-coeficient de variatie a presiunii dinamice de baza in raport cu inaltimea deasupra terenului liber gv-presiunea dinamica de baza stabilita la inaltimea de 10 m deasupra terenuluiObs. –se considera ca vantul produce suctiune si incarcarea prin actiunea vantului nu se ia in calcul

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STABILIREA TIPULUI DE INVELITOARE SI A GREUTATII PROPRII

gtic=ginv

c/cos α+gpic+pz

c=15,6/cos 2,31 +7+243,96=266,67 daN/m2

gti=380Kg/m2 -3,72 KN/m2=372 daN/m2

l=2,5 m -invelitoare de tip TCP/e -grosime δ=80mm +45 mm ginv

n1=14,4Kg/m2=14,4 daN/m2

ginvc1=ninv*ginv

n1=1,2*14,4=17,28 daN/m2

STABILIREA GREUTATII PROPRII A PANEI

gtcc=ginv

c1/cos α+gpnc+gcv

c+ge1c+gpi

c+pzc=17,28/cos

2,31+27,5+4,4+3,6+7+243,96=303.75 daN/m2

LT=t=10,5 m ;a=2,95 m gtcc=303,75 daN/m2

gpnn=21,5 daN/m2

gpnc=npn*gpn

c=1,1*21,5=23,65 daN/m2

gt3c=ginv

c/cos α +gpnc+gcv

c+gc1cgpi

c+pzc=17,28/cos

2,31+23,65+4,4+3,6+7+243,96=299.90~300 daN/m2

gt3n=ginv

n/cos α+gpnn+gcv

n+ge1n+gpi

n+pzc(n)=14,4/cos

2,31+21,5+4+3+5+158,5=205,95 daN/m2

STABILIREA INCARCARILOR PE MAL DE PANA

amed=a1/2+a2/2=2950/2+2950/2=2950mm=2,95mqt

c=qt3c*amed=300*2,95=885daN/m

qtn=qt3

n*amed=205,95*2,95=607,55daN/mqx

c=qtc*cos α=885*cos 2,31=884,28daN/m

qxn=qt

n*cos α=607,55*cos 2,31=607,06 daN/mqy

c=qtc*sin α=885*sin 2,31=35,67 daN/m

qyn=qt

n*sin α=607,55*sin 2,31=24,49 daN/m

Mxc1=qx

c*LT2/11=884,28*10,52/11=8862,9 daN/m

Mxr=Mx

c=Mxcq=qx

c*LT2/16=884,28*10,52/16=6093,24 daN/m

Mxc2=qx

c*LT2/32=884,28*10,52/32=3046,62 daN/m

Myc2=qy

c*(1,3*e2)2/8=35,67*(1,3*3,5)2/8=92,31 daN/m

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Dimensionarea si verificarea sectiunii panei de acoperisin camp curent

OL 37,2 →R=2200dan/cm2 t≤16mma) din conditia de rezistenta sect I σ=Mx/Wx+My/Wy≤1,1R/Wx

Kw=Wx/Wy Kw=7÷9 Kw=8÷10 σ=Mx+(Wx/Wy)My≤1,1R*Wx

- consideram profil Kw=9σ2=Mxc

2+Kw*Myc2≤1,1R*Wx

Wxnec≥(Mxc

2+Kw*Myc2)/1,1/R=(6093,24+9*92,31)*102/(1,1*2200)=286,12

cm2

J24-Wx=354 cm2

b)din conditia de rigiditatefx

2=0,151*(qxn*LT

n)/Jx≤fa=LT/200 Jxnec≥(0,151*qx

n*LTn)/fa

Jxnec≥(0,151*qx

n*LTn)/(LT/200)=( 0,151*607,06*10,54*10-2)/(1050/200)=

2122,3 cm4

J20-Jx=2140 cm4

-Dimensioarea rationala se face din conditia de rezistenta cu profilul J24 –alesJ24 : A=46,1 cm2 G=36,2 Kg/cm Jx=4250 cm4 Jy=221 cm4 h=240 mmWx=354 cm3 Wy=41,7 cm3 b=106 mmIx=9,59 cm Iy=2,20 cm d=8,7 mm

VERIFICARIa) verificari de rezistenta:

sect I σ=Mx/Wx+My/Wy≤1,1*R(2420 daN/cm2) σ=6093,24*102/354+92,31*102/41,7=1942,62 daN/cm2<1,1*R(2420 daN/cm2)

b) verificari de rigiditate

frez=(fx2+fy

2)1/2 ≤fa=LT/200fx= 0,151*qx

n*LTn/Jx=0,151*607,06*10-2*10,54/4250=2,62 cm

fy=5/384*qyn*l2

4/Jy=5/384*24,29*10-2*3504/(2,1*106*221)=0,10 cmfrez= = 2,622 cm <fa=LT/200=5,25 cm

verificari de zveltete

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λx=lfx/ix=0,5 *LT/ix=0,5*1050/9,59=54,74λy=lfy/iy=l2/jy=(LT/3)/iy=(1050/3)/2,20=159,09λmax=max(λx, λy) λmax≤λa(=250)λmax=max(54,74;159,09)=159,09<250

c) verificari din conditii constructive

J24>J20

CALCULUL PANEI DE ACOPERIS IN CAMPUL MARGINAL:

Mxc1=qx

c*LT2/11=8862,9daN/m

Mxr=Mx

c=Mxcq=6093,24daN/m

Myc=My

c1=Mymax=92,31daN/m

DIMENSIONAREA SI VERIFICAREA PANEIDE ACOPERIS IN CAMPUL MARGINAL

1) din conditia de rezistentaσ=M/W≤Rσ=Mx/Wx+My/Wy≤1,1R/Wx= σ2Mx+Wx/Wy≤1,1RWx

K= Wx/Wy → 7÷9 U →8÷10 J

- profile I Kw=9 σ=Mx+Kw*My≤1,1RWx σ=Mx

c1+KwMy≤1,1 RWx

Wxc1nec(σ)=Wxc1

nec*h/2=400,56*24/2=4806,79 cm4

2) din conditia de rigiditatefc1~0,307*qx

n*LTn/Jxc1≤fa≤LT/200

Jxc1nec(f)≥200*0,307*qx

n*LTn/LT=200*0,307*607,06*10-

2*10,54/1050=4314,87 cm4

Jxc1nec=max(Jc1

nec(σ);Jc1nec(f))=max(40806,79;4314,84 cm4) =Jc1

nec(σ)=4806,79 cm4

Jxcmp-Jx

nec-Jx=4086,79-4250=556,79 cm4

JxU=Jx

cmp/2=556,79/2=278,4 cm4=U12 –cu Jx=364 cm4

A=17 cm2 G=13,4 Kg/m

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h=120 mm b=55 mm d=7mmJx2=364 cm4 Jy2=43,2 cm4

Wx2=60,7 cm3 Wy2=11,1 cm3

Ix2=4,62 cm Iy2=1,59 cm l=1,60 cm

Jxc1=Jx1+2Jx2=4250+2*364=4978 cm4

Wxc1=Jx/(h/2)=4978/(24/2)=414,83 cm3

Ixc1= = =7,88 cm

Jyc1=Jy1+2*Jy2=221+2(43,2+17*(0,87/2+1,60)2)=448,2 cm4

Wyc1=Jy

c1/(bU+dI/2)=448,2/(5,5+0,87/2)=75,52 cm3

Iyc1= = =2,37 cm

Verificarea sectiunii adoptate

1) verificari de rezistentaσ=Mx

c1/Wxc1+My

c1/Wyc1≤1,1R

σ=8862,9*102/414,83+92,31*102/75,52=2258,75 daN/cm2 <1,1R(=2420 dan/cm2)

2)verificari de rigiditate frez= ≤fa=LT/200 fyc1=5/384*qy

n*l24/(G*Jy

c1)=5/384*24,49*10-2*3504/ (2,1*106*448.2)=0,05 cmfxc1=0,307*qx

n*LTn/Jx

c1=0,307*607,06*10-2*10,54/4978=4,55 cmfrez= ~4,55 cm<LT/200=5,25 cm

3) verificari de zveltete λx=lfx

c1/Jxc1=0,7*LT/Jx

c1=0,7*1050/7,88=93,27 λy=lfy

c1/Jyc1=l2/Jy

c1=1050/3/2,37=147,68 λmax ≤ λa=250 λmax=max(λx; λy)=max(93,27;147,68)=147,58<250

Stabilirea zonei de consolidarea zonei in campul marginal

4z(0,875LT-z)/(0,875LT)2*Mxc1=Mcap

q=0 z1,z2

LT=10,5 m Mx

c1=8862,9 daN/m4z(0,875LT-z)=(0,875LT)2*Mcap

2/Mxc1

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σ=M/W ≤R→Mcap2=Wx1*R=354*2200=7788 daN/m

-4z2+3,5zLT-(0,875LT)2Mcapc1/Mx

c1=0-4z2+3,5z*10,5-(0,875*10,5)2*7788/8862,9=0-4z2+36,75z-74,17=0 →z1=2,99 ~3 m z2=6,19 ~6,2 m z1

real=z1-300=3000-300=2700mm=2,7 m z2

real=z2+300=6200+300=6500 mm=6,5 m

Calculul tirantului

fortele axiale in tirantiNmax=max(Nn-1,Nn)

Nn-1= =(n-1,5)V Nn= =(n-0,5)V/2cosβ

V=1,25qye/2=1,25*35,67*10,5/2=234,08 daNcosβ==a2/ =3/ =0,65Nn=(5-0,5)*234,08/(2*0,65)=810,29 daNNn-1=(5-1,5)*234,08=819,08 daNNmax=819,28 daNd0≥ Rt=2000 daN/cm2

d0nec≥

Calea de rulareCalculul grinzii de rulare

I Stabilirea incarcaturilor A) Incarcari permanente

1° greutatea proprie a grinzii de rulare

ggrn=(15÷30)t=(15÷30)10,5=(157,5÷315) daN/m

ggrn=300 daN/m ngr=1,1

ggrc=ngr*ggr

n=1,1*300=330 daN/m 2° greutatea de proprie a sinei de rulare gs

n=(1,10÷1,20)b2

γ02=(1,1÷1,2)*0,092*7850=(69,94÷76,30) daN/m gs

n=70 daN/m ns=1,1 gs

c=ns*gsn=1,1*70=77 daN/m

3 ° greuatea TGS

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gTGSn=(1,10 ÷1,15)gTGS*bTGS/2=(1,1 ÷1,15)*42*1/2=(23,1 ÷24,15)

daN/m gTGS5=42 daN/m2

bTGS=bS/2-(20 ÷30)+c-b/2-L60÷100-(20 ÷40)= =370/2-(20 ÷30)+1000-90/2-(60 ÷100)-(20 ÷40) ~1000 mm=1 m gTGS

n=24 daN/m nTGS=1,1 gGTS

c=gTGSn*nTGS=24*1,1=26,4

B) Incarcari cvasipermanente gu

n150 daN/m2 *bTGS/2=150*1/2=75 daN/m nu=1,4 gu

c=gun*nu=75*1,4=105 daN/m

Total incarcari permanente +cvasipermanente

gTn=ggr

n+gsn+gTGS

n+gun=300+70+24+75=469 daN/m

gTc=ggr

c+gsc+gTGS

c+guc=330+77+26,4+105=538,4 daN/m

C) Incarcari variabile Actiunea podului rulant 1 ° lucrari verticale a)normele statice P1

n=22,5tf=22500 daN P3n=27tf=27000 daN

P2n=8tf=8000 daN P4

n=10,5tf=10500 daN A=6300 mm=6,3 m B=7900 mm=7,9 m

b) normale dinamice (Ψ) Pi

n Ψ’= Ψ’Pin

Ψ-coeficientul dinamic a fetei de grinda de functionare a P.R. GF –IV → Ψ=1,4 pt 2 P.R. Ψ’= Ψ-0,1 Ψ’=1,4-0,1=1,3 P1

n Ψ’= Ψ’P1n=1,3*22500=29250 daN

P2n Ψ’= Ψ’P2

n=1,3*8000=10400 daN P3

n Ψ’= Ψ’P3n=1,3*27000=35100 daN

P4n Ψ’= Ψ’P4

n=1,3*10500=13650 daN Pmax

nΨ=P3nΨ=P3

n* Ψ= Ψ*P3n=1,4*27000=37800 daN

de calcul dinamice Pi

n Ψ’=nV*PinΨ’ nV=1,2

P1cΨ’=nV*P1

nΨ’=1,2*29250=35100 daN P2

cΨ’=nV*P2 nΨ’=1,2*10400=12800 daN

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P3 cΨ’=nV*P3

nΨ’=1,2*35100=37800 daN P4

cΨ’=nV*P4 nΨ’=1,2*13650=16380 daN

PmaxcΨ=P3

cΨ=nV*P3 nΨ=1,2*37800=45360 daN

2 °lucrari (actiuni) orizontale a) produse de franare P.R. -normate statice Pe

n=f*Pmaxn=f*P3

n=0,1*27000=2700 daN f=0,1 -coeficient de frecare a rotilor pe sina

-normate dinamice PC

nα=α*Pln=1,5*2700=4050 daN

α=1,5 (GF – IV ,Qc=400KN) -de calcul dinamice Pl

cα=n0*Pl nα=1,3*4050=5265 daN ;n0=1,3

α cα=2*Pl cα=2*5265=10530 daN

b)produse de franarea caruciorului -normate staticeHn=f*(Qn+Gc

n)/nrc*nrct=0,1*(45000+7000)/4*2=2600 daN

nrc-numar roti caruciornrc

t-numar roti franate Pt

n=Hn/nrp=2600/2=1300 daN nrp-numar roti PR pe aceeasi cale Pt

nα= α*Ptn=1,5*1300=1950 daN -normate dinamice

Pt cα=n0*Pt

nα=1,3*1950=2535 daN -de calcul dinamic

Stabilirea schemei statice

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-grinda simplu rezemata a)actiuni permanente +cvasipermanente P+CVP A=6,3 m B=7,9 m

b)actiuni variabile produse de PR 1)verticale -cazuri posibile

A+B-A+A=6,3+7,9-6,3+6,3=14,2>t →3 forte

=0 →P1(1,6*9,5)+P3*7,9=R*x=35100(1,6+9,5)+37800*7,9 =145800 R=P3+2P1+P3=37800+2*35100+37800=145800 →x=4,72 m d=7,9-4,72=3,18 m P1=P1

cΨ’=35100 daN P3=P3

cΨ’=37800 daN

=0 →VF*10,5-P1(10,5-2,06)-P3(10,5-2,06-1,6)-P10,54=0 VF=1/10,5(35100*8,44+37800*6,84+35100*0,54) VF=54642,86 daN =0 →VA*10,5-P1(10,5-0,54)-P3(10,5-0,54-6,3)-P1*2,06=0 VA=1/10,5(35100*9,96+37800*3,66+35100*2,06) → VA=53357,14 daN

η2/ η1=(10,5-1,6)/10,5) → η2=0,848η3

/ η1=(10,5-1,6-6,3)/10,5 → η3=0,248η4/ η1=(10,5-1,6-6,3-1,6)/10,5 → η4=0,093Tmax max

cΨ’= =P3 cΨ’* η1+P1

cΨ’* η2+P3 cΨ’* η3+P1

cΨ’* η4

=37800(1+0,248)+35100*(0,848+0,095) =80273,7 daNMmax max

pq=Mmax maxp cΨ’+Mmax

qtc=143832,87+7419,86=151252,73 daN/mTmax max

pq=Tmax max cΨ+Tmax

qtc=80273,7+2826,6=83100,3 daNMmax max

ptcα=Mmax max cΨ’* / =

143832,7*3*2535/(2*35100+37800)=10128,23 daN/m

II Stabilirea dimensiunilor sectiunii transversale cu verificarea sectiunii

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stabilirea dimensiunilor inimii a)inaltimea inimii -conditii constructive 1)h=(1/8 ÷ 1/10)t=(1/8 ÷ 1/10)*10,5=(1,31 ÷1,05) m -conditia de rigiditate 2)nmin=5/24 *R*t2/(E*fa*nv)=5/24*2200*10502/(2,1*1061,75*1,2) nmin=114,58 cm R=2200 daN/cm2

t=10,5 m E=2,1*106 daN/cm2

ta=t/600=1050/600=1,75 ;nv=1,2

3)conditii de rezistenta hopt=1,15 Wx

nec ≥Mmax maxpq/Rred=151252,73*102/1980=7639 cm3

Rred ~(0,85 ÷0,95)R=(0,85 ÷0,95)*2200=(1870 ÷2090) daN/cm2

hmin=114,58 ~115 cm ;t>6 mm

ti(cm) 0,6 0,7 0,8 0,9 1,0 1,2hiopt(cm) 139 128 120 114 108 98λ=hi/ti 231 183 150 127 108 82

hiopt=1,23*hopt=1,10* =1,15* h1opt=113,24 cm h2opt=120,34 cmhi-rot 50 mm<1000 mm hi=120 cm=1200 mm 100 mm<1000 mm ti=0,9 cm=9 mm

4)gosimea inimii a)conditii constructive ti ≥6 mm b)conditii de rezistenta ti ≥1,185 Tmax max

pq/(hi*Rf)=83100,3/(120*1300)=0,599 cm=5,99 mm

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Rf=1300 daN/cm2

c)conditii de apasare a rotilor

z=50+2(hs+t)=50+2*(90+20)=270 mm=27 cm σl=P3

cΨ/(z*ti) ≤R →ti ≥P3 cΨ/(z*R)=45360/(27*2200)=0,76 cm

t=0,8 cm (tabel) Stabilirea dimensiunilor talpilor a)conditia de stabilitate locala b’/t ≤k1 → b’ ≤k1*t k1=15 (OL37) bi’ ≤2*b’=2ki*t Anec ≥2,73* Wx

nec=7639 cm3 λet=1200/9=133 Anec ≥2,73* =207,45 cm2 →Anec=210 cm2

Anec=Ai+Ata’pi

Ai=hi*ti=120*0,9=108 cm2 →Ata’pi=210-108=102 cm2

A1t=Ata’pi/2=102/2=51 cm2

A1t=t*bi=t*2*k1*t=2*k1*t2 →t= = =1,30 cm=13 mm

t>6 mm t=13 mm=1,3 cm bi ≤2*b’=2k1*t=15*13=195 mm →b’=200 mm=20 cm b’ ≤k1t bi ≤2b’=2k1t=2*20=40 cm Ata’pi=2*bi*t=2*40*1,3=104 cm2

Ai=bi*ti=120*0,9=108 cm2

a)b3

b2=b3net+2d=bi+2d=400+2*20=440 mm =44 cm

b3net=bi=400 mm=40 cm

O=20 mm =2 cm b)b3 ≥hs+2*(bc+(20 ÷40)+(10 ÷15)) =90+2*(80+(20 ÷40)+(10 ÷15))=(310 ÷360) mm bc=80 mm- α 80*80*8 hs=90 mm

Calculul grinzii de rulare

15

Stabilirea incarcaturilor a)Permanente -greutate tabla striata gTGS

n=gTGS*l/2=51/2=25,5 daN/m gTGS

c=gTGSn*n=25,5*1,1=28,1 daN/m n=1,1

-greutate profil U16 ggf

n=18,8 Kg/m =18,8 daN/m n=1,1 ggf

c=ggfn*n=18,8*1,1=20,7 daN/m

b)cvasipermanente -incarcare utila gu

n=150....200 daN/m*l/2=200/2=100 daN/m n=1,4 gu

c=gun*n=100*1,4=140daN/m

Total incarcari -gtgf

n=144,3 daN/m =gtgf

c=188,8 daN/m

Tmaxc=gtgf

c/2*l=188,8*10,5/2=991,2 daNMmax

c=gtgfc*l2/8=188,8*10,52/8=2601,9 daN/m

STABILITATEA INCARCARILOR

-din conditia de rezistenta σ=M/W ≤Rred →Wx

nec=Wmaxc/Rred

Rred=0,85R(=2200 daN/m) ≥2601,9*102/1870=139,14 cm3

U18: Wx=150 cm3

-din conditia de rigiditatefef=5/48*Mmax

n*l2/(E*Jxnec) →Jx

nec ≥Mmaxn*l2/(E*fa)*5/48

Mmaxn=ggtf

n*e2/8=144,3*10,52/8=1988,63 daN/mfa=l/250=1050/250=4,2 cmJx

nec=5/48*1988,63*102*10502/(2,1*106*4,2) →Jxnec=2589,36 cm4

U22 Jx=2690 cm4

Wx=245 cm3

Verificarea ansamblului GR-GF

16

in punctul A σA

1= σAp+ σA

L+ σAML ≤R

σA2= σA

P+ σAPt ≤bR

σAP=Mmax max

p,q/Wxnet

Wxnet=Jx

net/ymax

Jxnet=40*12263/12-2*19,55*1203/12=512170,59 cm4

Wxnet=512170,59/61,3=8355,15 cm3

Agrnet=1,3*40+120*0,9+1,3*44-2*1,3*2,1=211,74 cm2

Mmax maxpq=151252,73 daN/m

σAP=151252,73*102/8355,15=1810,29 daN/cm2

σALL cα/Agr

net=10530/211,74=49,73 daN/cm2

σAML=L cα*e/Wx

net=10530*70,3/8355,15=88,6 daN/cm2

l=hgr/2+b=1226/2+90=703 mm=70,3 m σA

1=1810,29+49,73+88,6=1948,62 daN/cm2 ≤R(=2200daN/cm2) σA

pt=Mmax maxqtcα/Wxf

A

VGF= / =((0,5*95)*57,5+37,4*117,86)/(0,5*95+1,3*44,0+37,4)JyF=(1,3*443/12+1,3*44*50,242)+(0,5*953/12+0,5*95*19,762) +(197+37,4*67,622)=379082,22 cm4

Wyft=Jy

t/xA=5415,5 cm3

Pt cα=2535 daN

Mmax maxpecα=10128,23 daN/m=10128,23*102daN/cm

σApt=101128,23*102/5415,5=187,02 daN/cm2

σA2=1810,29+187,02=1997,31 daN/cm2<1,1R=2420 daN/cm2

-in punctul C ≤Rf=1540 daN/cm2

=Tmax maxpq*Sx

1/2/(ti*Jx net) Sx

1/2=A*yG=40*1,3*(120*1,3)/2+120/2*0,9*120/2=6393,8 cm3

=83100*6393,8/(0,9*512170,59)=1152,66 daN/cm2<Rf=1540 daN/cm2

- in punctul Dσech

D= ≤mRσD

x= σDP+ σD

L+ σDML

σDP= σA

P*hi/h=1810,29*120/122,6=1771,9 daN/cm2

σDL= σA

L=49,73 daN/cm2

σDML= σA

ML*hi/h=88,6*120/122,6=86,72σD

x=1771,9+49,73+86,72=1908,35 daN/cm2

= *Sxt/Sx

1/2=1152,66*3736,04/6393,8=673,52 daN/cm2

17

Sxt=1,4*44*(120+1,3)/2=3736,04 cm3

σDe=Pmax/(z*ti)=45360/(28*0,9)=1800 daN/cm2

σDech= =219

2 daN/cm2

σDech=2192 daN/cm2<mR=1,1*2200=2420 daN/cm2

Verificarea de rigiditate fx=5/48*Mmax max

pc/nmed*l2/(E*Ixbr) ≤fa=e/600=1,75 cm

nmed=1,3 Mmax max

p cΨ=151252,73*102 daN/cm JG=(40*1,3*1,3/2+120*0,9*(120/2+1,3)+44*1,3(120+1,3+1,3/2))/ (40*1,3+120*0,9+44*1,3)=62,91 cm Jx

br=40*1,33/12+40*1,3*62,912+0,7*1203/12+0,9*120*1,612+44*1,33/12 +44*1,3*59,042

Jxbr=506277,34 cm4

tx=5/48*151252,73*102*10502/(1,3*2,1*106*506277,34)=1,26 cm<1,75 cm

Calculul imbinarii talpii superioare cu inima

3mm ≤as ≤0,7*tmin=0,7*9=6,3 mm

Verificare la stabilitate generala

l1 ≤Kg*Iyf Kg=40 Iyf= = =51,65 cm l1=1050<<40*51,65 =2066 cm -stabilitatea generala este asigurata

Verificarea la stabilitate locala a elementelor grinzii de rulare

a)stabilitatea locala a inimii hi/ti=120/0,9=133,33 →hi/ti>80* 80* =78,2 -sunt necesare rigidizari curente pe grinda

a ≤2*hi=2*120=240 cm 1050/240=4,37 -5 panouri 1050/5=210 cm br=hi/n0+40 mm=1200/30+40=80 mm br ≤(bs-ti)/2=(440-9)/2=215,5 mm

18

br ≤(bi-ti)/2=(400-9)/2=195,5 mm →br=100 mm tr ≤br/15=100/15= 6,67 mm -otet RC=1,5R=3300 daN/cm2

-rigidizarea de reazem Anec

str ≥Rmaxgr/Rc=83100/3300=2518 cm2

-br1=380 mm ≤bi →tr1=25,18/38=0,66 cm →tr1=0,7 cm tr1=7 mm = otet lat 380*7 →AR1

ef=26,6 cm2

15*ti* =15*0,7* =10,25 cm

Verificarea la strivire a rigiditatii de reazem

Astref=br1*tr1=38*0,7=26,6 cm2

σ2ef=83100/26,6=3124,06 daN/cm2 <3300 daN/cm2

Verificarea la compresiune cu flambaj

Tmax maxp,ct ≤R

Ar11=Ar1

ef+15*ti* *ti26,6+10,25*0,7=33,78 cm2 λy1=efy1/iy1=120/10,31=11,36 efy1=hi120 cm ey1=0,993 → 83100/(0,993*33,78)=2477,37<2200 daN/cm2

-consideram : 420*8=3360 mm2=33,6 cm2

Ar1’=Ar1ef+15*t1 *ti=33,6+10,25*0,8=41,8 cm2

83100/(0,993*41,8)=2002 daN/cm2≤R=2200 daN/cm2

Calculul sudurilor dintre rigidareade reazem si inima

3mm≤as1≤0,7*tmin

=Rmaxgr/(2*Rf

s*es) →as≥83100/(2*1540*120)=0,22 →as1=3 mm

Verificarea la stabilitate locala a inimii

hi/tr≥70 →120/0,8=150 →necesara verificarea ≤n=0,9

19

→ 35100*3,6+37800*5,2=VB*10,5 → VB=30754 daN M1=42146*3,0=126438 daN/m M2=42146*4,2-35100*0,6=155953 daN/m Mmed=(M1+M2)/2=(126438+155953)/2=141196 daN/m Tcoresp=46146 daN σD=Mmed/Ws

WxD=Jx

br/ys=506277,34/60=8437,96 cm3

σD=141196*102/8437,96=1673,34 daN/cm2

=Tcoresp/(ti*hi)=46146/(0,9*120)=427,28 daN/cm2

σcr=K0*(ti/hi)2*102 (KN/cm2) σecr=K1(ti/a)2*102 (KN/cm2) =(1250+950/α2)*(ti/a)2*102 (KN/cm2) α=2100/1200=1,75 γ =c*b0/hi*(t/ti)3=2*44/120*(13/9)3=2,21 σe

D=Pmax/(z*ti)=43360/(28*0,9)=1720,6 daN/cm2

σe/ σ=1720,6/1673,34=1,02 →K0=7028,35 a/hi=1,75>0,8 → σecr=K1(ti/a’)2

σcr=7028,35*(0,9/120)2*102=39,53 KN/cm2=3950 daN/cm2

σecr=9385*(0,9/210)2*102=52,79 KN/cm2=5279 daN/cm2

=(1250+950/1,752)*(0,9/210)2*102=8,77 KN/cm2=877 daN/cm2

=0,89<0,9 (=n)

Verificarea la oboseala a grinzii de rulareσmax

pq=Mmax max/Wxnet ≤ γ*R

-intindere γ0=c/(b-a*l) a=1,55 b=0,95 c=0,95 P1=22500 daN P3=27000 daNMmax max

pnqn=Mmax maxpcqc*1/nmed

=151252,73*1/1,3=116348,25 daN/mσmax

pnqn=116348,25*102/8355,15=1392,53 daN/cm2

σmin=Mmaxqn/Wmax

net=7419,86*102/(1,3*8355,15)=68,31 daN/cm2 →ρ=0,0491γ=0,95/(0,95-1,55*0,0491)=1,09σmax1392,53< γR=1,09*2200=2391,4 daN/cm2

Calculul grinzii cu zabrele

20

Stabilirea incarcarilorNormate Coef. inc. De calcul

A)Permanente-din invelitoare (tip

TPC)ginv

n=13 daN/m2

-din panegpn

n=1,1*np*gI*t/(t*L)=1,15*10*3,62*10,5

/(10,5*105)=3,96 daN/m2

-greutate luminatorge

n=5 daN/m2

-greutate covgcov

n=4,5 daN/m2

-greutate proprie fermaL=30 mt=10,5 m

gt2=303,75 daN/m2

gfn=19,5 daN/m2

1,2

1,1

1,1

1,1

1,1

ginvc=15,6 daN/m2

gpnc=4,36 daN/m2

gec=5,5 daN/m2

gcovc=4,95 daN/m2

gfc=21,45 daN/m2

gtn=45,96 daN/m2 gt

c=51,86 daN/m2

B)cvasipermanente-greutate instalatii

gen=3 daN/m2

-din praf industrialgpi

n=5 daN/m2

1,4

1,4

gec=4,2 daN/m2

gpic=f daN/m2

C)variabile

-zapada : cf STAS 10101/21-92 Pz

n=Ce*Cz”*gz=0,8*1,1*150=120 daN/m2 -calculat la pana γpz= γa-0,4*gt

n/(Ce*gz)=2,2-0,445,96/(0,8*150)=2,047>0,3* γa=0,66 (pt SLR) γ0=2,2 γc=1,4 -pt clasa de importanta III γ0z= γc-0,2*gt

n/(Ce*gz)=1,4-0,2*45,96/(0,8*150)=1,323>0,42 (pt SLEN) Pz

c=Pzn* γFz=120*2,047=245,64 daN/m2

Pzn(q)=Pz

n* γ0z=120*1,323=158,76 daN/m2

-din actiunea vantului Pv

n=β*cvi*ch(z)*gv

H=S+300÷400+hpana+hinv

21

=18700+350+240+70=19360 mm H’=H+440+h0

=19360+440=19800 mm β=1,6 –coeficient de rafala pt. constructii din categoria C1

gv=0,41 KN/m2 –presiunea de baza stabilizata, la inaltimea de 10 m deasupra terenului

Cvi-coeficient aerodinamic pe suprafata “i”ch(z)-coef de variatie a presiunii de baza in raport cu inaltimea z

deasupra terenului liber

β=45°-cf STAS 10101/20 -90H/R=19360/30000=0,65 →Cv1=-0,62α=2,31H’/R=19800/30000=0,66 →Cv1’=0,305α=45°H/L=0,65L/l=105000/30000=3,5 →Cv3=-0,53H/L=0,65 →Cv2=-043H’/l=0,66 →Cv2’-0,432 -amplasament de tip II ch(z)=0,65*(z/10)0,44≥0,65γF1= γ0=1,2 -SLURγav= γc=1 -SLEN pt. clasa de importanta III si constructii (C1)

z1=(H’-H)/2+H=19,58 m=z2

z1’=H’+he/2=19,8+3/2=21,3=z2’ ch(z1)=0,65*(19,58/10)0,44=0,874>0,65 ch(z1’)=0,65*(21,3/10)0,44=0,907>0,65Pv1

n=β*Cv1*ch(z1)*gV=1,6*(-0,62)*0,874*42=-36,41 daN/m2

Pv1n’= β*Cv1’*ch(z1’)*gV=1,6*0,305*0,907*42=-18,59 daN/m2

Pv2n= β*Cv2*ch(z2)*gV=1,6*(-0,43*0,874*42=-25,26 daN/m2

Pv2n’= β*Cv2’*ch(z2’)*gV=1,6*(=0,432)*0,907*42=-26,33 daN/m2

Pv0n= β*Cv0*chmed*gv=1,6*0,8*0,703*42=37,79 daN/m2

Pv3n= β*Cv3*chmed*gv=1,6*(=0,53)*0,703*42=-25,4 daN/m2

-valori de calculat ale incarcarilor din vant(SLUR) Pvi

c= γFv*Pviω - γFv=1,2

Pv1c=1,2*(-36,41)=-43,69 daN/m2

Pv1’c=1,2*18,59=22,31 daN/m2

22

Pv2c=1,2*(-15,16)=-30,31 daN/m2 →asupra acoperisului

Pv2’c=1,2*(-26,33)=-31,60 daN/m2

Pv0c=1,2*37,79=45,35 daN/m2

Pv3c=1,2*(-25,4)=-30,48 daN/m2 →asupra peretilor halei

-valori normate ale incarcarilor din vant (SLEN) -verificarea la rigiditatePvi

n(c1)= γov*Pvin ; γov=1 →Pvi

n(c1)=Pv1’n

-incarcari pe ferma: *permanente si cvasipermanente qp

n=(ginvn/cosα+gpana

n+gen+gcv

n+gpn+ge

n+gpin)*t

=(13/cos2,31 +3,96+5+4,5+19,5+3+5)*10,5=556,7 daN/m qp

c=(ginvc/cosα+gpana

c+gcc+gcv

c+gtc+ge

c+gpic)*t

=(15,6/cos 2,31++4,36+5,5+4,95+21,45+4,2+7)*10,5=662,8 daN/m *zapada qz

n=Pzn(c1)*t=158,76*10,5=1667 daN/m

qzc=Pz

c*t=245,64*10,5=2580 daN/m *vantqvi

n=Pvin*t qvi

c=Pvic*t

qv1n=-36,41*10,5=-328,3 daN/m qv1

c=-43,69*10,5=-458,75 daN/mqv1’

n=18,59*10,5=195,2 daN/m qv1’c=22,31*10,5=234,3 daN/m

qv2n=-25,26*10,5=-265,2 daN/m qv2

c=-30,31*10,5=-318,3 daN/mqv2’

n=-26,33*10,5=-276,5 daN/m qv2’c=-31,00*10,5=-331,8 daN/m

qv0n=37,79*10,5/2=198,4 daN/m qvo

c=45,35*10,5/2=238,1 daN/mqv3

n=-25,4*10,5/2=-133,35 daN/m qv3c=-30,48*10,5/2=-160 daN/m

Eforturi axiale in barele grinzii cu zabrele-incarcari P+CV qP+CV

n=566,7 daN/m qP+CVc=662,8 daN/m

P1=qP+CV*3/2=850 daNP2=qP+CV*(3+3)/2=1700 daNV= /2=(2*P1+9P2)/2=(2*850+1700*9)/2=8500 daN

-incarcari z qz

n=1667 daN/m P1=qz*3/2=1667*3/2=2500 daN P2=qz*(3+3)/2=1667*(3+3)/2=5000 daN

23

V=(2*P1+9*P2)/2=(2*2500+9*5000)/2=25000 daN -incarcarea =100 daN –la mijlocul deschiderii V= /2=100/2=50 daN -incarcarea 1=100 daN -in reazeme

Calculul static al cadrului transversal

Incarcari P+CV din reactiunea grinzii cu zabrele Rgz

P+CV(n)=qP+CVn*α/2=566,7*30/2=8500 daN

RgzP+CV(c)=qP+CV

c*α/2=662,8*30/2=9942 daN din greutate stalp gs

s=(100.....150)=125 daN/m Gs

n*hs=125*6,5=812,5 daN Gsc=812,5*1,1=893,75 daN

gsi=(200....250)=225 daN/m

Gin=gs

i*hi=225*12,2=2745 daN Gic=2745*1,1=3019,5 daN

Schema statica

l=bi/2-bs/2=1200/2-370/2=415 mm din greutate pereti: gp

n~ginvn=13 daN/m2*t/2=13*10,5/2=68,25 daN/m

gpc=gp

n*1,1=68,25*1,1=75,1 daN/m Gp

s(n)=gpn*hs=68,25*6,5=443,6 daN

Gps(c)=gp

c*hs=75,1*6,5=488,15daN →partea superioara Gp

i(n)=gpn*hi=68,25*12,5=832,65 daN

Gpi(c)=gp

c*hi=75,1*12,5=916,22 daN →partea inferioara

- calculul reactiunii la nivelul consolei inferioare din incarcari P+CVRP+CV

n=RgzP+CV(n)+Gs

(n)+Gps(n)=8500+812,5+443,6=9756,1 daN

RP+CVc=Rgz

P+CV(s)+Gs(c)+Gp

s(c)=9942+893,75+488,15=11324 daN nP+CV=11324/9756,1=1,16Mmax

n(P+CV)=Rn*e=9756,1*0,415=4048,78 daNmMmax

c(P+CV)=Rc*e=11324*0,415=4700 daNm incarcari variabile

din reactiunea grinzii cu zabrele provenite din incarcarea din zapadaVgz

z(n)=qzn*L/2=1167*30/2=25005 daN

Vgzz(c)=qz

c*L/2=2580*30/2=38700 daNnz=38700/25005=1,55

calculul reactiunii din nivelul consolei inferioare din zapada

24

Rz(n)=Vgzz(n)=25005 daN

Mmaxz(n)=Rz(n)*e=25005*0,415=10377 daNm

Rz(c)=Vgzz(c)=38700 daN

Mmaxz(c)=Rz(c)*e=38700*0,415=16060,4 daNm

din actiunea podurilor rulantea)verticale -reactiuni maxime si minime

η1=1η2=1/10500*4200=0,4η3=1/10500*(6300+2600)=0,848η4=1/10500*2600=0,248Rmax=P1(η2+ η3)+P3(η1+ η4)Rmin=P2(η2+ η3)+P4(η1+ η4) -valori normate P1

nΨ=29250 daN P3nΨ=35100 daN

P2nΨ=10400 daN P4

nΨ=13650 daNRmax

n=29500*(0,4+0,848)+35100*(1+0,248)=80308,8 daNRmin

n=10400*(0,4+0,848)+13650*(1+0,248)=30014,4 daN -valori de calcul P1

cΨ=35100 daN P3cΨ=37800 daN

P2cΨ=12800 daN Pc

cΨ=16380 daNRmax

c=35100*(0,4+0,848)+37800*(1+0,248)=90979,2 daNRmin

c=12800*(0,4+0,848)+16380*(1+0,248)=36416,6 daN nprV=90979,2/80308,8=1,1-calculul momentelor maxime si minime la nivelul consolei inferioare provenite din excentritatea grinzii pe stalp

l’=c-l=1000-415=585 mm=0,585 m-normate Mmax

n=Rmaxn*l’=80308,8*0,585=46980 daNm

Mminn=Rmin

n*l’=30014,4*0,585=17558 daNm-de calcul Mmax

c=Rmaxc*l’=90979,2*0,585=53223 daNm

Mminc=Rmin

c*l’=36416,6*0,585=21304 daNm

b)orizontale

Rmaxpt=Pt*Σηi

Ptn=1950 daN

Ptc=2535 daN

RmaxPt(n)=1950*(1+0,4+0,848+0,248)=4867,2 daN

25

RmaxPt(c)=2535*(1+0,4+0,848+0,248)=6327,36 daN

npro=6327,36/4867,2=1,3

x=hs-hgr=6500-1226=5274 mm=5,274 m din actiune vantului

α=2,31β=45°

-normateR1=qV1*l1=382,3*11,815=4516,87 daN R1

V=4516,87*cosα=4513,2 daN R1

O=4516,87*sinα=182,1 daNl1= =11815 mm=11,815 mR1’=qV1’*e1=195,2+4,24=827,65 daN R1

o’=827,65*sinβ=585,24 daN R1

V’=827,65*cosβ=585,24 daNl1’= =4,24 mR2’=qV2’*l2’=276,5*4,24=1172,36 daN R2

O’=3133,34*sinβ=828,98 daN R2

V’=1172,36*cosβ=828,98 daNR2=qV2*l2=265,2*11,815=3133,34 daN R2

o=3133,34*sinα=126,3 daN R2

V=3133,34*cosα=3190,8 daN

Calculul reactiunilor transmise de grinda cu zabrele stalpului, provenite di actiunea de vant

ΣMA=0 → VB*30-R2V*24,1-R2

V’*16,5+R1V’*13,5-R1

V*5,9- (R1o-R2

o) *0,295+(R1

o’+R2o’)*1,795=0 → VB=3511,18 daN

ΣMB=0 →VA*30-R1V*24,1+R1

V’*16,5+R2V’*13,5-R2

V*5,9+(R1o-R2

o)*0,295 (R1

o’-R2o’)*1.795=0 →VA=4376,56 daN

ΣPx →HH=R1o’+R2

o’+R2o+R1

o=1358,42 daN

RV=4376,56 daN qVo=198,4 daN/mMV=4376,56*0,415=1816,27 daN/m qV3=135,33 daN/m

Calculul static al cadrului transversal tinandcont de influenta conlucrarii spatiale

I1/I2=1/5÷1/10=1/7

26

I2=7*I1 n=I2/I1=7a=hs=6500 mmH=18700 mm λ=a/H=6500/18700=0,348α1=((n-1)* λ+1)/n=((7-1)*0,348+1)/7=0,441α2=((n-1)* λ2+1)/n=((7-1)*0,3482+1)/7=0,247α3=((n-1)* λ3+1)/n=((7-1)*0,3483+1)/7=0,179α4=((n-1)* λ4+1)/n=((7-1)*0,3484+1)/7=0,155

Incarcari permanente + cvasipermanente

MP+CV=4048,78 daN m Ψ*H=6,5 → Ψ= λ=M/H*3/(2* α3)*( α2- Ψ2)=4048,78/18,7*3/(2*0,179)*(0,247-0,3482)

=228,42 daNMA= *H-M=228,42*18,7-4048,78=222,67 daN m

ZAPADA

Mmaxz(n)=10377 daN m

z=M/H*3/(2* α3)*(0,247-0,3482)=585,44 daNMA= *H-M=585,44*18,7-10377=570,68 daN m

PODURI RULANTEa)verticaleMmax=46980 daN mMmin=17668 daN m

PRV=3*(α2-λ2)/(H*2*α3*2)*(Mmax+Mmin)=3*(0,247-0,3482)/ (2*18,7*2*0,179)*(46980+17558)=3641,02 daN/2=1820,51 daN

b)orizontaleΨ*H=x →Ψ=x/H=5,274/18,7=0,282Rmax

Pt=4867,2 daNpro=Rmax

Pt*[1-Ψ/(2*α3)*(3*α2-Ψ2)] =4867,2*[1-0,282/(2*0,179)*(3*0,247-0,2822)]=2331,14 daN

A=H( pro-RmaxPt*(1-Ψ))=18,7*(2331,14-4867,2*(1-0,282))= -

2157,63daNm

VANT -asupra acoperisuluiMV=1816,27 daN mHa=1358,42 daN

27

Va=Ha/2+M/H*3/(2*α3)*( α2-λ2) =1358,42/2+1816,27/18,7*3/(2*0,179)*(0,247-0,3482)=781,68 daN

-asupra paretilorqVo=198,4 daN/m qVs=135,33 daN/m

Vp=3*H/16*α4/α3*(qVo-qVs)=191,49 daN

in barele grinzilor cu zabrele cosiderand gr de cadruzapada in. din act. vantului inc. din poduri rulante

asupra acoperisului

asupra peretilor

act. verticale act. orizontale

element

bara + - + - + - + - + -

1 2 13 14 15 16 17 18 19 20 21 22Ts 1-2 174

83- 23370 - - 5708 - 108811 69661 -

2-3 18281

- 24437,5 - - 5969 - 119781 72844 -

3-4 38025

- 50830 - - 12415 - 236665 151515

-

4-5 38306

- 51205 - - 12507 - 238413 152264

-

5-6 44279

- 59190 - - 14457 - 275587 176433

-

Ti 7-8 - 28928 - 38670 9445 - 180047

- - 115268

8-9 - 41526 - 55510 13558 - 258456

- - 165466

9-10 - 41734 - 55788 13625 - 259749

- - 166294

Δ 1-7 - 22616 - 30232 7384 - 140761

- - 90116

3-7 15962

- 21337 - - 5211 - 99345 63601 -

5-8 5692

- 7609 - - 1584 - 35427 22681 -

3-8 - 12256 - 16383 4001 - 76279 - - 48834

5-10 - 2867 - 3832 936 - 17841 - - 11422

M 2-7 1240

- 1658 - - 405 - 7719 4942 -

4-8 1152

- 154` - - 376 - 7173 4592 -

5-9 - 1199 - 1603 392 - 7564 - - 47796-10 989 - 1322 - - 323 - 6153 3939 -

eforturi axiale in barele gr. cu gr. simplu rezemata efortuei axiale

28

zabreleperm.+cvasiperm zapada inc. conc.

unit. in reazem

inc.conc. unit. in mijlocul

desc.

perm. cvasiperm.

elment

bara

+ - + - + - + - + -

1 2 3 4 5 6 7 8 9 10 11 12Ts 1-2 - 106

55- 25719 - 2988,

5- 2946 6814 -

2-3 - 11109

- 26787 - 3125 - 3086 7125 -

3-4 - 23235

- 55882 - 6500 - 6558 14820 -

4-5 - 23422

- 56339 - 6548 - 6609 14929 -

5-6 - 26969

- 64781 - 7569 - 7716 17257 -

Ti 7-8 18341

- 44325

- 4945 - 5034 - - 11275

8-9 25724

- 61846

- 7098,5

- 7298 - - 16185

9-10

25841

- 62120

- 7134 - 7337 - - 16266

Δ 1-7 14311

- 34570

- 3866 - 3943 - - 8814

3-7 - 10317

- 25041 - 2728,5

- 1796 6221 -

3-8 6756

- 15930

- 2035 - 2089 - - 4777

5-8 - 3428,5

- 8193 - 973 - 1031 2218 -

5-10

1648

- 3890

- 490 - 553 - - 1117

M 2-7 - 1806

- 4895 - 212 - 215 483 -

4-8 - 1788

- 4872 - 197 - 200 449 -

29

5-9 192,5

- 172 - 205 - 204 - - 467

6-10

- 1645

- 4506 - 169 - 260 385 -

Gruprarea incarcarilorNormale De calcul

Pn+CVn+ng*Σvi

n Pc+Cvc+ngΣvi

c

+ - + -

Ts 1-2 - 20318 - 258042-3 - 31861 - 404633-4 - 66548 - 845154-5 - 67449 - 856605-6 - 77143 - 97971

Ti 7-8 52922 - 67211 -8-9 73761 - 93677 -9-10 74086 - 94089 -

Δ 1-7 41273 - 52417 -3-7 - 20544 - 260903-8 18842 - 23930 -5-8 - 9757 - 123915-10 4623 - 4623 -

M 2-7 - 8968,5 - 113904-8 - 8761 - 111275-9 128 - 162 -6-10 - 5519 - 7009

x’max=x’ijP+CV+x’ij

xP+CV+0,9*(x’ijgz+x’ij

xz+x’ijXprv+x’ij

pro)

Dimensionarea barelor grinzii cu zabrele

A)Bare intinse I Talpa inferioara Nmax=94089 daN -otel OL37 R=Ri=Rc=2200 daN/cm2 →t≤16 mm1. Alcatuirea sectiunii2. Stabilirea dimensiunilor gguseului tg=f: →tg=8....10 mm →tg=10 mm3. Stabilirea dimensiunilor sectiunii a)din conditia de rezistenta

30

σ=x’/Anet≤R →Abrnec≥Nmax/R*Ks

Ks=1,1÷1,2 -coeficient de slabire Ks=1,15 Abr

nec≥94089/2200*1,15=49,18 cm2

A1necbr=Abr

nec/2=49,18/2=24,59 cm2 →L 120*120*12 L→ 2*L →

b=2*R+1=25 cm Iy= =5,35 cmIy=(Jy+A1*d2)*2=(368+27,5*3,92)*2=1572,55 cm4

d=l+tg/2=3,40+1/2=3,9 cm

4)verificarea sectiunii a)verificarea de rezistentaσmin=Nmax/Anet ≤R=94089/49,96=1883,28daN<R(=2200daN/cm2)Anet=Abr-asc=55-5,04 cm2=49,96 cm2

Asc=2*(2,1+1,2)=5,04 cm2

b)verificarea de zvelteteλx=lfx/ix ≤ λa=400 →300/3,66=81,96< λa

λy=lfy/iy ≤ λa →lfynec ≥iy* λa=5,35*400=2140

Lgr/2=3000/2=1500 →l1=1500 cm λy=1500/5,35=280,37< λa

DIAGONALA INTINSA -bara 1-7 N=52417 daNl17=3,97 m=3970 mmlfx=0,8*l1-7=0,8*3970=3176 mmlfy=l17=3970 mm

1. ALCATUIREA SECTIUNII

2. Stabilirea dimensiunilora)conditia de rezistenta Ks=1 -imbinari sudateAbr

nec ≥Nmax/R*Ks=52417/2200*1=23,89 cm2 →Abrnec=23,83/2=11,92 cm2

L70*70*10 2*L

3. VERIFICAREA SECTIUNII a)de rezistenta σmax-Nmax/Anet=52417/24,1=2175 daN/cm2<R(=2200 daN/cm2) b)de zveltete: λx=lfx/Px=317,6/2,1=151< λa=400

31

λy=lfy/Iy=39,70/3,33=191< λa

Solidarizari

l1 ≤80*i1

l1 ≤80*3,66=292,8 →293 cm -pt talpa inferioaral1<80*2,1=168 cm -pt. diagonala intinsa -Cel putin o fisura la barele intinse

B)Dimensionarea barelor comprimate Talpa superioara Nmax=N5-6=97971 daNl5-6=3000 mm →lfx=e5-6=3000 mm lfy=e5-6=3000 mm R=2200 daN/cm2

1)Alcatuirea sectiuni

2)Dimensionarea sectiunii Metoda coeficientilor de profilKx=4,6Ky=1,9 tg=12 mm → de tipul sectiunii (tabel)αx=0,3 αy=0,2

ξx=lfx* =300* =96,42ξy=lfy* =300* =61,97

→ →

emin=min(ex;ey)=0,678hnec ≥lfx/( αx* λx)=300/(0,3*77)=12,98 cmbnec ≥lfy/( αy* λy)=300/(0,2*59)=25,42 cmh1nec ≥129,8 mmb1

nec≥(bnec-tg)/2=(25,42-1,2)/2=12,1 cm=12 mm →L130*130*12

Metoda coeficientilor de flambaj impus initialφ°=0,57Anec

br≥N5-6/2=97971/0,57*2200=78,13 cm2

A1nec=78,13/2=39,07 cm2 →L150*150*14 A1

ef=40,3 cm2

Ix1=Iy1=845 cm4 Aef=2*40,3=80,6 cm2

ix1=iy1=4,58 cm →2*L Ix=2*Ix1=1690 cm4

l=4,21 cm Ix=[845+40,3*(4,21+1,2/2)2]*2=3555cm4

ix=4,58 cm

32

iy=6,64 cm

3.Verificarea sectiunii a) de rezistenta σ=N/Anet=97971/75,56=1297 daN/cm2<<R=2200 daN/cm2

Anet>φmin*As2=0,57*75,56=43,,07 cm2 –verificarea de rezistenta nu este necesarab)verificarea de stabilitate N5-6/(lmin*As2)=97971/(0,57*80,6)=2132,5 daN/cm2<R=220 daN/cm2

DIAGONALA COMPRIMATANmax=N3-7=26090daNl37=4175 mm=417,5 cmlfx=0,8*l37=334 cm=3340 mm R=2200 daN/cm2

lfy=417,5=4175 mm

1.Alcatuirea sectiunii

2.Dimensionarea sectiunii ξx=lfx= =334*

=208ξy=lfy* =417,5* =167

→ →

hnec≥lfx/(αx*λx)=334/(0,38*129)=8,63 cmbnec≥lfy/(αy*λy)=417,5/(0,2*112)=18,64 cmh1nec>8,63 cm=86,3 mmb1nec≥1864/2=932 mm →L100*100*10

φmin=min(φx; φy)=0,38Anec≥N37/(φ0*R)=26090/(0,38*2200)=31,21 cm2

A1nec≥31,21/2=15,61 cm2 →L100*100*10

L 2*L

3.Verificarea sectiunii a)de zveltete λmax=110<λa=150 pt. diagonale b)de stabilitate N34/(lmin*Aef)=26090/(0,38*34,2)=2007,5 daN/cm2<R(2200 daN/cm2)

Montantii comprimati

33

lfx=0,8*2800=2240 mmlfy=2800 mm

1. Alcatuirea sectiunii

2. Dimensionarea sectiunii →ξx=lfx* =224*

=211ξy=lfy* =280* =170

→lx=0,37 λx=127 →ly=0,46 λy=115

hnec≥lfx/(αx*λx)=224/(0,3*127)=5,88 cm=58,8 mmbnec≥lfy/(αy*λy)=2800/(0,2*115)=12,12 cm=121,2 mmb1nec=121,2/2-12/2=60 mm →L60*60

lmin=0,37Anec≥N2-7/(φ0*R)=11390/(0,37*2200)=13,99 cm2

A1nec=Anec/2=7 cm2 →L70*70*6

L70*70*6 2*L

3.Verificarea sectiunii a)de zveltete λx=lfx/ix=224/2,28=98 λy=lfy/iy=280/3,40=82 →λmax=98<λa=150 b)de stabilitate N2-7/(φmin*Aef)=11390/(0,37*15)=2052,25 daN/cm2<R(=2200 daN/cm2)

Solidarizari

l1≤40*i1

l1≤40*4,58=183 cm -pt. talpa superioaral1≤40*3,04=122 cm -pt. diagonala comprimatal1≤40*2,28=91 cm -pt. montanti

-Se vor excuta cel putin 2 tururi la barele comprimate

34

Verificarea la rigiditate a grinzii cu zabrele

fef=ΣNikn* *lik/(G*Aik) ≤fa

fa=L/300=3000/300=10 cmelement bara Nik nik lik E Aik Nik*nik*lik/(G*Aik)

Ts 1-2 -20318 -2988,5 295 80,6 0,10482-3 -31861 -3125 295 80,6 0,17353-4 -66548 -6500 295 80,6 0,75394-5 -67449 -6548 295 80,6 0,76985-6 -77143 -7569 300 80,6 1,0349

Ti 7-8 52922 4945 590 55 1,33688-9 73761 7098,5 295 2,1*106 55 1,33739-10 74086 7134 300 55 1,3728

Δ 1-7 41273 3866 397 26,2 1,15133-7 -20544 -2728,5 417,5 38,4 0,29023-8 18842 2005 414 26,2 0,29705-8 -9757 -973 435 38,4 0,05125-10 4623 490 435 26,2 0,0179

M 2-7 -8968,5 -212 280,7 16,26 0,01624-8 -8761 -197 305 16,26 0,01545-9 128 205 317,5 16,6 0,00246-10 -5519 -169 315 16,26 0,0804

Pe jumatate de grinda 8,8028Pe toata grinda 17,6056

fef=17,6 cm ≥fa

Calculul prinderii zabrelelor in noduri

Nodul 2N72=11390 daNbc=70 mmtc=6 mmtg=12 mml=1,93 cm =19,3 mm

35

l1=bc-l=70-19,3=50,7 mmimpunem 3 mm ≤as1 ≤min(0,7*tg=8,4 mm;0,85*tc=5,1 mm) →as1=4 mm 3 mm ≤as2 ≤0,7*tmin=6 mm →as2=4 mmDin conditia de rezistenta ls1 ≥N2-7*l1/(2*as1*L0c*Rf

s)=11990*5,07/(2*0,4*7*1500)=6,87 cm →ls1=70 mmRf

s=0,7*R=0,7*2100=1500 daN/cm2 -sudura de coltls2 ≥N2-7*l/(2*as2*bc*Rf

s)=11390*1,93/(2*0,4*7*1500)=2,61 cm →ls2=30 mm

*Din conditii constructive ls1,2 ≥40 mm ls1,2 ≥bc=70 mm 15*as1=6<ls1 ≤60*as1=24 15*as2=6<ls2 ≤60*as2=24 →ls1=70 mm ls2=70 mm

ls1r=ls1+2*as1=70+2*4=78 →ls1

r=80 mm ls2

r=ls2+2*as2=70+2*4=78 →ls2r=80 mm

Nodul 3 -N3-7=26090 daN -diagonala comprimata -N3-8=23930 daN -diagonala intinsa

Calculul prinderii diagonalei 3-7

N3-7=26090 daN bc=100 mm as1=6 mm tc=10 mm 3 mm ≤as1 ≤min(0,7*tg=8,4 mm; tg=12 mm 0,85*tc=8,5 mm) l=28,2 mm as2=6 mm l1=bc-l=100-28,2=71,8 mm 3 mm ≤as2 ≤0,7*tmin=8,4 mm

*Din conditia de rezistenta ls1 ≥26090*7,18/(2*0,6*10*1500)=10,41 cm →105 mm ls2 ≥26090*2,82/(2*0,6*10*1500)=4,09 cm →45 mm

*Din conditii constructive ls1,2 ≥20 mm ls1,2 ≥bc=100 mm ls1,2=9 cm<ls1,2 ≤60*as1,2=36 cm → ls1=105 mm

36

ls2=90 mm ls1

r=l1+2*as1=105+2*6=117 mm →ls1r=120 mm

ls2r=l2+2*as2=90+2*6=102 mm →ls2

r=105 mm

Calculul prinderii diagonalei 3-8

*Din conditia de rezistenta ls1 ≥23930*4,91/(2*0,6*7*1500)=9,33 cm →ls1=95 mm ls2 ≥23930*2,09/(2*0,6*7*1500)=3,97 cm →ls2=40 mm

*Din conditii constructive ls1,2 ≥40 mm ls1,2 ≥bc=70 mm 18*as1,2=3 cm ≤ls1,2 ≤60*as1,2=36 cm →ls1=95 mm ls2=70 mm ls1

r=95+2*0,6=96,2 mm →ls1r=100 mm

ls2r=70+2*0,6=71,2 mm →ls2

r=75 mm

Nodul 8

Calculul prinderii diagonalei 8-5

as1=6 mm3 mm ≤as1 ≤min(0,7*tg;0,85*tc)=8,4 mmas2=6 mm3 mm ≤as2 ≤t min -7 mm

*Conditia de rezistenta ls1 ≥12391*7,18/(2*0,6*10*1500)=4,94 cm=49,4 mm →ls1=50 mm ls2 ≥12391*2,82/(2*0,6*10*1500)=1,94 cm =19,4 mm →ls2=20 mm

*Conditii constructive ls1,2 ≥40 mm ls1,2 ≥bc=100 mm 9 ≤ls1,2 ≤36 cm →ls1=ls2=100 mmls1

r=ls2r=100+2*0,6=101,2 mm=105 mm

Calculul montantului 4-8 N4-8=1127 daN bc=70 mm tc=6 mm tg=12 mm

37

l=1,93 cm=19,3 mm l1=70-19,3=50,7 mm as1=5 mm 3 mm≤as1 ≤min(0,7*tg;0,85*tc)=5,1 mm as2=4 mm 3 mm ≤as2 ≤0,7*tmin=4,2 mm

*Conditia de rezistenta ls1 ≥11127*5,07/(2*0,5*7*1500)=5,37 cm →ls1=55 mm ls2 ≥11127*1,93/(2*0,4*7*1500)=2,56 cm →ls2=30 mm

*Conditii constructive ls1,2 ≥40 mm ls1,2 ≥bc=70 mm 7,5 ≤ls1 ≤30 cm 6 ≤ls2 ≤24 cm →ls1=70 mm ls2=70 mm

Nodul 1Calculul prinderii barei 1-7 N1-7=52417 daN bc=70 mm tc=10 mm tg=12 mm l=20,3 mm l1=70-20,3=49,7 mm as1=6 mm 3 mm ≤as1 ≤(0,7*tg;0,85*tc)=8,4 mm as2=6 mm 3 mm ≤as2 ≤0,7*tmin=7 mm

*Din conditia de rezistenta ls1 ≥N1-7*l1/(2*as1*bc*Rf

s)=52417 daN*4,97/(2*0,6*7*1500)=20,7 cm=210 mm ls2 ≥ N1-7*l/(2*as2*bc*Rf

s)=52417*2,03/(2*0,6*7*1500)=8,44 cm=85 mm

*Din conditii constructive ls1,2 ≥40 mm ls1,2 ≥bc=70 mm 15*as1,2=9 cm ≤ls1,2 ≤60*as1=36 cm →ls1=210 mm ls2=85 mm

38

ls1r=210+2*6=222 mm →ls1

r=225 mm ls2

r=85+2*6=97 mm →ls2r=100 mm

Calculul stalpului

Incarcari Permanente

in pct. B NB=9942+(75,1+138)*5,274=11066 daN MB=245*5,274=1292,13 daN m TB=245 daN Variabile

MB=907*0,274=4784 daN m

Din reactiunea verticala a podului rulant

MB=2003*5,274=10564 daN m -din reactiunea orizontala a podului rulant -din vant asupra acoperisului -din vant asupra peretilor

Scema Partea sect. ef. sect. inc. perm. inc. zapada react. V a P.R(Rmax)

0 1 2 3 4 5 6superioara A M 0 0 0

N -9942 -38700 0T 245 907 -2003

B M 1292 4784 -10564N -11066 -38700 0T 245 907 -2003

Csup M 1592,5 5895,5 -13020N -1327 -38700 0

39

T 245 907 -2003inferiaora Cinf M -3107,5 -10164,5 40203

N -11327 -38700 -90980T 245 907 -2003

D M -118,5 900 15767N -15269 -38700 -90980T 245 907 -2003

React. H a PR Vant asupra acop.

Vant asupra pereti

gr. inc. Mmax Ncoresp=Nmax

st. dr. st. dr.7 8 9 10 11 12 13 140 0 0 0 0 0 -11207 daN m -44775 daN0 0 5252 0 -43390 -43390

3030 -3030 480 -230 2429 -363115980 -

159802215 2097 15804 -16156

0 0 5252 0 -44514 -445143030 -3030 480 1025 3684 -237611938 -

119382730 3533 12669 -11207

0 0 5252 0 -44775 -447753297 -3297 480 1317 4243 -235111938 -

119384910 3533 47312 23436 87819,5

daNm-139697 daN

0 0 5252 0 -134855

-134855

3297 -3297 480 1317 4243 -235128285 -

282855674 37312 87819,5 31249,5

0 0 5252 0 -139697

-139697

3297 -3297 480 4421 7294 700

Stabilirea dimensiunilor sectiunii transversale

P2=43390 daNP1=90980 daNIs/Ji=1/7=0,143 →Ji=7*Is

hs/hi=6,5/12,2=0,533

1°Stabilirea lungimii de flambaj a)pentru ramura inferioara a1)in planul cadrului efx=μ1*hi=1,386*12,2=16,9 mμ1= = =1,386C=(P2+P1)/P2=(43390+90980)/43390=3,097μ12=1,94μ11=1,02 f(hs/hi=0,533;Is/Ji=0,143)

40

a2)in plan perpendicular cadrului efy

i=hi=0,2 m b)pentru ramura superioara b1)in planul cadrului efx

s= μ2*hs=1,73*6,5=11,25 m μ2= μ1/C1=1,386/0,8=1,73 C1=hs/hi* =6,5/12,2* =0,8 b2)in plan transversal efy

s=hs=6,5 m

Dimensionarea ramurii superioare a cadrului

Nmax=44775 daNMmax=11207 daNm -compresiune cu flamba -incovoiere in jurul axei xti=(1/60÷1/75)*h=6,67÷5,35 mm →ti=6 mm

Kx=0,435Ky=1,4 αx=0,4....0,45=0,44αy=0,22....0,26=0,25 →functie de tipul reactiuniiξx=lfx* =11,25* =164,5ξy=lfy* =650* =170

→lxo=0,494

λxo=117

→lyo=0,464

λyo=118

bnec≥lfy/(αy* λy)=650/(0,25*118)=22,03 cmhnec ≥lfx/( αx* λx)=1125/(0,44*117)=21,85 cmlmin

o=lyo=0,464

Anec ≥Nmax/(lmino*Rredus)=44775/(0,464*1870)=51,60 cm2=5160 mm2

Rredus=0,85*R=1870 daN/cm2

Vom considera ti=6 mm b=250 mm t=10 mm

41

hi=h-2*t=380 mm A=72,8 cm2

Jx=21860 cm4

Wx=1093 cm3

Ix=17,3 cm Jy=2600 cm4

Wy=208 cm3

Iy=6 cm

Verificarea dimensiunilor i)de zveltete λx=lfx/Ix=1125/17,3=65 λy=lfy/Iy=650/6=108 → λmax=108< λa

ii)de rezistenta σ=Nmax/A+Mcoresp/Wx=44775/72,8+11207*102/1093=1640 daN/cm2<R iii)verificarea stabilitatii generale Nmax/(lmin*Aef)+Cx*Mcoresp/(lg*(1- σ/ σxcr)*Wx)≤Rφmin=f(λmax,t,OL37)λmax=108 φmin=0,576lg=f(λtr,B,OL37)λtr=γ*λy

γ=f[(l/h)2;Jr/Jy]Jr= α ‘/s*Σ(h*t3+hi*ti

3)=1,2/3*(25,13*2+38*0,63)=23,28 cm4

α’=1,2(l/h)2*Jr/Jy=(650/402)*23,28/2600=2,36 → γ=0,721-incarcarea actioneaza la axa neutra μ=1 λtr=0,721*108=77,27 →lg=0,786σ=Nmax/Aef=44775/72,8=615,04 daN/cm2

Ncr*=π2*G*Ix/lfx

2= π2*2,1*106*21860/11252=357984,4 daNσ*cr=Ncr

*/A=257984,4/72,8=4917,37 daN/cm2

Cx=0,8544775/(0,575*72,8)+0,85*11207/(*,786*)_1-615,04/4917,37)*1093)=1158,3 daN/cm2<R(2200 daN/cm2) iv)verificarea la stabilitatea locala talpi b’/t≤K1 b’=(b-ti)/2=25-0,6=12,2 cm 12,2/1=12,2<K1

-inima Ψ= (σ- σ’)/ σ

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σ=Nmax/A+Mxcoresp/Ix*hi/2 =44775/72,8+11207*102/21860*38/2=1589,11 daN/cm2

σ’ =Nmax/A-Mx/Ix*hi/2=-359,03 daN/cm2

Ψ=(1589,11+359,03)/1589,11=1,23>0,5 →33=Ks

τ=T/A=4243/72,8=58,28 daN/cm2

β=0,07* τ/ σ*Ks=0,07*58,28/1589,11*33=0,0847h0/t=100* =14,36<38/0,6=63,3h0/t –valoarea maxima a inimii→seactiunea activa considerata C=15*ti* =15*0,6* =8,80 cm

A’=72,8-20,4*0,6=60,56 cm2

Jx’=Jx=0,6*10,22/12=21806,9 cm4

Wx’=Jx’*2/h=1147,7 cm3

Ix’= =18,98 cmJy’=Jy-10,2*0,63/12=2599,8 cm4

Wy=Jy*2/b=207,99 cm3

Iy’= =6,55 cmλx’=1125/18,98=59,27 λy’=650/6,55=99,23 φ=0,756 -verificarea de stabilitateσ=44775/(0,786*60,56)+0,85*11207*102/(0,786*(1-615,05/497,37)*1147,7)=2195 daN/cm2<R(2200 daN/cm2)h0=380 mm=38 cmh0<70*ti* =70*0,6* =41 cm → -nu sunt necesare rigidizari transversale pe inima stalpului

Dimensionarea ramurei inferioare a stalpului

Nmax=139697 daNMmax=87819,5 daNmlfx=16,9 mlfy=12,2 m

N2=N2M+N2

N

N2N=Nmax/2=139697/2=69848,4 daN

N2M=Mmax/bi=87819,5/1,2=73183 daN

N2=143031,5 daNN1

N=N1N-N1

M

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N1N=N2

N N1M=N2

M

N1N=-3334,5 daN

α=35°-50°l1=(hi-(a1+a2)/n)hi=12200 mma2=350 -inaltimea traversei de la baza stalpuluia1=400 -inaltimea traversei de la treapta stalpuluin=10 -nr. panouri

l1=(12200-(350+400))/10=145 mm

α=arctg 1145/1200=44 °є(35 °÷50 °)lc’=1659 mm

Dimensionarea ramurei 2)

Ky=0,48 (h>20)Kx=11 (h>30)αy=0,4αx=0,208 -lungimii de flambaj lfy=hi=12200=1220 cm lf1=l1=1145 mm=114,5 cm ξ1=lf1* =114,5* =47,1 ξy=lfy* =1220* =96,23

→

→

hnec≥lfy/( αy* λy)=1220/(0,4*80)=38,125bnec≥lf1/(α1*λ1)=114,5/(0,208*45)=12,24 →J40

Anec≥N2/(φmino*2)=143031,5/(0,694*2200)=98,68 cm2

h=40 cmb=15,5 cmd=1,44 cmA=118 cm2

Jy=29210 cm4

Wy=1460 cm3

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iy=15,7 cmJx=1160 cm4

Wx=149 cm3

ix=3,13 cm2

sy=857 cm3

Verificarea ramurei inferioare *verificarea de zveltete λx=lfx1/ix1=114,5/3,13=36,58 λy=lfy/iy=1220/15,7=77,7 →λmax=77,7<λa=120 *de stabilitate N2/(φmin*Aef) ≤R=143031,5/(0,804*118)=1507,6 daN/cm2<R(2200 daN/cm2) φmin=f(λmax=λy=77, 7,3,OL37) →φmin=0,804

Verificarea ramurei inferioare

Aef=2*118=236 cm2

Jx=2*[Jx1+A1*(b/2)2]=2*[1160+118*(120/2)2]=851920 cm4

Jy=2*Jy1=2*29210=58420 cm4

ix= = =60,08 cmiy= = =15,75 cm

*verificarea de zveltete λy=lfy’/iy’=1220/15,75=77,46 λx=lfx’/ix=16,9*102/60,08=28,13 →<λa=120 λx= λ1

2=u*A/Ad -zabrele confectionate din cornier L60*60*6 Ad=6,91 cm2

u=7 (sistem de imprastire, α) α=44° u=π2/(sinα*cos2α)=27,46 λ1

2=27,46*236/(2*6,91)=468,88 λxtr= =35,5<αa=120

Verificarea de rezistenta Nmax/Aef+Mmax/Wx≤R

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143031,5/236+87819,5*102/14198,67=1224,57 daN/cm2

Verificarea la stabilitatea generala Nmax/φmin*A+Cx*Mx/[(1-σ/σE)*Wx] ≤R φmin=f(λmax=λy=77,46 ;OL37,B) →φ=0,712 Cx=0,7 σE=3496 daN/cm2 (λ=77) σ=N/A=143031,5/236=606,06 daN/cm2

143031,5/(0,712*236)+0,7*87819,5*102/[(1-606,06/3496)*14198,67]=1374,97 daN/cm2≤R=2200 daN/cm2

Calculul zabrelelor Tmax

r=700 daN TC=0,012*A*R=0,012*236*2200=6230,4 daN T=Tr+TC=700+6230,4=6930,4 daN Δ=T/(2*cosα)=6930,4/(2*cos44)=4817,2 daN

lf=ld=1659 mm=165,9 cmL60*60*6 A=6,91 cm2

i=1,82 cmλ=lf/i=165,9/1,82=91,15≤ λa=150curba C -OL37 →φ=0,554 σ=4817,2/(0,554*6,91)=1258,36daN/cm2<o,7*R=1650 daN/cm2

Calculul sudurilor diagonalelor

l=1,69tc=6 mmti=21,6 cmΔ=4817,2 daN3 mm≤as1≤0,85*tc=5,1 mm → as1=3 mm3 mm≤as2≤0,7*tmin=0,7*6=4,2 mm →as2=3 mm

Din conditia de rezistenta ls1≥ Δ*l/(as1*bc*Rf

s)=4817,2*4,31/(0,3*6*1500)=7,68 cm=76,8 mm ls2 ≥ Δ*l1/(as2*bc*Rf

s)=4817,2*1,69/(0,3*6*1500)=3,02 cm=30,2 mm

Din conditii constructive15*as1=4,5 ≤ls1 ≤60*as1=18 cm →ls1=7,68 cm

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15*as2=4,5 ≤ls2 ≤60*as2=18 cm →ls2=3,02 →ls2

r=4,5 cm=45 mm ls1

r=ls1+2*as1=7,68+2*0,3=8,28 cm=8,5 cm=85 mm

Calculul montantilor

M=T/2=6930,4/2=3465,2 daNlm~bi=1200 mmlf=lm=120 cmA=5,63 cm2

I=1,5 cml=1,45 cm-cornier L50*50*6

→ φ=0,508

Verificarea la stabilitate σ=3465,2/(0,509*6,69)=1196,46 daN/cm2<R=1750 daN/cm2

Verificarea la zveltete λa=120> λ=100

Calculul sudurii 3 mm≤as1≤0,85*tc=5,1 mm →as1=3 mm 3 mm≤as2≤0,7*tc=4,2 mm →as2=3 mmls1≥3465,2*3,55/(o,3*6*1500)=4,55+2*as1=5,16 mm →ls1

r=55 mmls2≥3465,2*1,45/(0,3*6*1500)=2,1 mm →ls2

r=45 mm

Capitelul stalpului

VGP=VpA=43390 daN

Calculul sudurilor ce transmit eforturile de la placa 1) la placa 2) 3 mm≤as1≤0,7*tmin=0,7*6=4,2 mm

as1=4 mm tmin=min(tp,t1)=6 mm τs1=VGP/(4*as1*ls1) ≤Rf

s →ls1≥VGP/(4*ls1*Rfs)

Rfs=1500 daN/cm2

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Calculul sudurilor ce transmit eforturile de la diafragma 2) la inima stalpului

3 mm≤as2≤0,7*tmin=4,2 mm →as2=4 mm tmin=min(ti;tr)=6 mm τs2=VGP/(4*0,4*1500)=18,07+2*0,3=18,88 cm →ls2

r=19 cm =190 mm hr=ls2

r+20 mm=190+20=210 mm

Calculul treptei stalpuluiCalculul sudurii S3

3 mm≤as3≤0,7*tmin=7 →as3=6 mmτs3=N1/(as3*ls3) ≤Rf

s →ls3≥N1/(2*as3*Rfs)

N1=Ns/2+Ms/S3

Ns=Ncs=44775 daNMs=Mcs=12669 daNmN1=44775/2+12669/0,4=54060 daNls3≥54060/(2*0,6*1500)=30,03 cm+2*0,6=31,23 →ls3=32 cm

Calculul traversei

3 mm≤as4≤0,7*tmin=11,2 mmtmin=min(t=16 mm;tI40=21,6 mm)as4=10 mmN2=Ni/2+Mi/bi

N2=134885/2+47312/1,2=106854 daNNi=Nci=134855 daNMi=Mci=47312 daNmls4≥106854/(2*1*1500)=35,61+2*as4=37,61 cm →ls4

r=38 cm -sudurile S4 din interior se realizeaza constructiv -in calcul se considera doar cele din exterior

Tt=134855/2=67427,5 daNτt=Tt/A2=67427,5/76=887,2 daN/cm2<Rf

s=1500 daN/cm2

A2=38*2*1=76 cm2

Calculul bazei stalpului

ND=139697 daNMD=17819,5 daNmNS=ND/2+MD/bi=143031 daN

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Determinarea dimensiunilor in plan a placii de baza

a)dimensiunea in plan A*B A=h2+2*(tt+C1)=400+2*(15*60)=550 mm h2=400 tt=15 C1≥ht=4*15=60 mm Anec

st≥ND/Rbc=139697/70=1995,6 cm2

Rbc=70 daN/cm2 (c10=C8/10)

Bnec≥Anecst/A=1995,6/55=36,28 cm B≥bi+2*C2=120+2*20=160 cm

bi=120 cm bI=155 mm=15,5 cm C2=200 mm C2>b2=15,5 cm

Placa tip 1)

Mmax1=Rs*C1

2/2=70*62/2=1260 daNcm

Placa tip 3)l1/l2=200/400=0,5 →α3=0,06Mmax

3= α3*Rs*l1=0,06*70*202=1680 daNcm

Placa tip 4)l1/l2=400/300=1,33 α1=0,051 α2=0,0473M1= α1*Rs*l2

2=0,051*70*302=3213 daNcmM2= α2*Rs*l1

2=0,0473*70*402

Mmax4=5297,6 daNcm

Mmax=max(Mmax1,Mmax

3,Mmax4)=5297,6 daNcm

Wp=tp2/6 →Mmax/Wp ≤R

tp ≥ = =3,8 cm=38 mm →tp=38 mm

Determinarea inaltimii traversei

3 mm ≤as6 ≤0,7*tmin=7 mm tmin=min(tt1=10 mm,tI40=21,6 mm) as6=7 mm τs6=Ns/(4*as6*ls6) ≤Rf

s →ls3 ≥Ns/(4*as3*Rfs)

ls6 ≥143031/(4*0,7*1500)=34,06 cm+2*as3=35,5 cm

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ls6r=36 cm

MI-I=Rs*A/2*lI2=70*55/2*6,92=91649 daNcm

II-I=Rs*A/2*l2=70*55/2*6,9=13282,5 daNτ=II-I/At=13282,5/1,36=368,96 daN/cm2<Rf

s=1500 daN/cm2

σ=MI-I/Wt=91469/259,2=352,89 daN/cm2<R=2200 daN/cm2

Wt=tt*ht2/6=1,2*362/6=259,2 cm3

Verificarea sudurii S7

τs7=N3/(as7*Σls7) ≤Rf

s

3 mm<as7 ≤0,7*tmin=7 mm tmin=min(tp=38 mm, tt=10 mm)=10 mm →as7=5 mm Σls7=2*b-2*as7=2*1600-1*5=3190 mm =319 cm τs7143031/(0,5*319)=896,5 daN/cm2≤Rf

s=1500 daN/cm2

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