1. Grafichkia Rad - AB STUB

1) POS S −ABSTUB

−AB ploča

d pl=λ

35=510

35=14,57 cm=¿usvajamd pl=16cm

−ABgreda

bg=30cm ;dg=L

12=600

12=50cmb /d=30/50

−analiza opterećenjameđuspratna konstrukcija

−parket : gp=0,024∗8,0=0,192 kN /m2

−blindit :gbl=0,025∗18,0=0,450kN /m2

−mršavi beton : gmb=0,040∗22,0=0,880 kN /m2

−MSK : gmsk=0,160∗25,0=4,000 kN /m2

−malter : gmal=0,015∗19,0=0,285kN /m2

∑ gs=5,807 kN /m2; ps=1,50kN /m2

−krovna MSK−cementna košuljica : gck=0,03∗22,0=0,660 kN /m2

−cementnimalter : gcm=0,015∗19,0=0,285 kN /m2

−sloj za pad : gszp=0,03∗22,0=0,660 kN /m2

−termoizolaci ja :g ti=0,05∗1,0=0,050 kN /m2

−mršavi beton : gmb=0,04∗22,0=0,880 kN /m2

−MSK : gmsk=0,16∗25,0=4,000 kN /m2

− plafon: gpl=0,015∗19,0=0,285kN /m2

∑ gk=6,820 kN /m2; pk=1,50 kN /m2

sk=0,75 kN /m2

−sopstvenatežina gredeggr=0,30∗0,50∗25,0=3,75 kN /m

−ukupno svedenoopterećenje (MSK )gsu=gs∗λ+ggr=5,807∗5,1+3,75=33,37 kN /mpsu=ps∗λ=1,5∗5,1=7,65kN /m

−ukupno svedenoopterećenje (krovna konstrukcija−prohodna)gku=gk∗λ+ggr=6,82∗5,1+3,75=38,53 kN /mpku=pk∗λ+sk∗λ=(1,5+0,75 )∗5,1=11,48 kN /m

−dimenzionisanje stubova

N gmax=(38,53+3∗33,37 )∗6,0=831,34 kN

N pmax=(11,48+3∗7,65 )∗6,0=206,58 kN

N qmax=N g

max+N pmax=1038,42kN

σ0

0,7∗f bk≤0,35

σ 0

0,7∗35=0,35=¿σ0=8575,0 kN /m2

σ 0=N q

max

Ab=¿8575,0=1038,42

Ab=¿ Ab=0,121m2

−usvajamkvadratni presek stuba=¿0,121=a2=¿a=34,79 cm=¿USVAJAM a=35cm

−greda:A x=0,3∗0,5=0,15m2

I z=0,3∗0,53

12=0,003125m4

ggr=0,30∗0,50∗25,0=3,75 kN /m

−stub :A x=0,35∗0,35=0,1225m2

I z=0,35∗0,353

12=0,001251m4

ggr=0,35∗0,35∗25,0=3,0625kN /m

−šemaopterećenja

−Seizmičkemasem=100 %∗g+50%∗p

m1=(33,37∗24,0+5∗4,0∗3,0625 )+(0,5∗7,65∗24,0 )=953,93 kN

m2=(33,37∗24,0+5∗3,0∗3,0625 )+ (0,5∗7,65∗24,0 )=938,61 kN

m3=(33,37∗24,0+5∗3,0∗3,0625 )+ (0,5∗7,65∗24,0 )=938,61 kN

m4=(38,53∗24,0+5∗1,5∗3,0625 )+ (0,5∗11,48∗24,0 )=1085,45 kN

SASS IZLAZNI FAJLSTRUCTURE POS AB RAM - SEIZMICKE MASETYPE PLANE FRAMENUMB OF JOIN 25NUMB OF MEMB 36NUMB OF SUPP 5NUMB OF LOAD 4TABU ALLJOIN COOR1 0. 0. SUPP UNTIL 5 STEP 1 6. 0.6 0. 5. UNTIL 10 STEP 1 6. 0.11 0. 8. UNTIL 15 STEP 1 6. 0.16 0. 11. UNTIL 20 STEP 1 6. 0.21 0. 14. UNTIL 25 STEP 1 6. 0.MEMB INCI1 1 6 UNTIL 56 6 11 UNTIL 1011 11 16 UNTIL 1516 16 21 UNTIL 2021 6 7 UNTIL 2425 11 12 UNTIL 2829 16 17 UNTIL 3233 21 22 UNTIL 36MEMB PROP PRIS1 THRU 20 AX 0.1225 IZ 0.00125121 THRU 36 AX 0.1500 IZ 0.003125CONS E 32500000. ALLLOAD 1 * SEIZMICKE MASE*JOIN LOAD8 FORC X 953.9313 FORC X 938.6118 FORC X 938.6123 FORC X 1085.45TRACESOLVEPROBLEM CORRECTLY SPECIFIED, EXECUTION TO PROCEED

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STRUCTURE POS AB RAM - SEIZMICKE MASE

========================================================================

REZULTATI PRORACUNA SEIZMICKIH SILAPREMA TEORIJI DINAMIKE KONSTRUKCIJA

BROJ ETAZA N= 4 BROJ RAMOVA M= 1

MATRICA FLEKSIBILNOSTI:

.59208E-04 .63497E-04 .63878E-04 .63946E-04

.63497E-04 .85405E-04 .88320E-04 .88668E-04

.63878E-04 .88320E-04 .10880E-03 .11172E-03

.63946E-04 .88668E-04 .11172E-03 .13252E-03

SOPSTVENE VREDNOSTI

.34366E-01 .27301E-02 .77848E-03 .38516E-03

SOPSTVENI VEKTORI

TON OSCILOVANJA: 1.35544 .47073 .54867 .59248

TON OSCILOVANJA: 2-.66760 -.45167 .09819 .58366

TON OSCILOVANJA: 3.58708 -.38424 -.57771 .41707

TON OSCILOVANJA: 4-.33379 .67747 -.62150 .20822

KOEFICIJENAT OBJEKTA KO= 1.00000

KATEGORIJA TLA - PODLOGE = 2

KOEFICIJENAT DUKTILITETA KP= 1.00000

KOEFICIJENAT SEIZMICNOSTI KC= .050

SEIZMICKE SILE - TON OSCILOVANJA: 1

KOEFICIJENAT DINAMICNOSTI KD= .60097

PERIODA OSCILOVANJA T= 1.1648

SPRAT AMPLITUDE SPRATNE CVOR SEIZMICKEOSCILOVANJA TEZINE SILE1 .5999 954. ( 8) 20.5802 .7945 939. ( 13) 26.8173 .9261 939. ( 18) 31.2574 1.0000 1085. ( 23) 39.034


KOEFICIJENAT DINAMICNOSTI KD= 1.00000

PERIODA OSCILOVANJA T= .3283

SPRAT AMPLITUDE SPRATNE CVOR SEIZMICKEOSCILOVANJA TEZINE SILE1 -1.1438 954. ( 8) -64.3192 -.7739 939. ( 13) -42.8163 .1682 939. ( 18) 9.3084 1.0000 1085. ( 23) 63.984




SPRAT AMPLITUDE SPRATNE CVOR SEIZMICKEOSCILOVANJA TEZINE SILE1 1.4076 954. ( 8) 56.5612 -.9213 939. ( 13) -36.4243 -1.3851 939. ( 18) -54.7644 1.0000 1085. ( 23) 45.722




SPRAT AMPLITUDE SPRATNE CVOR SEIZMICKEOSCILOVANJA TEZINE SILE1 -1.6030 954. ( 8) -32.1582 3.2536 939. ( 13) 64.2213 -2.9847 939. ( 18) -58.9154 1.0000 1085. ( 23) 22.827

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STRUCTURE POS AB RAM - SEIZMICKE MASE===================================================================TON OSCILOVANJA: 1 UTICAJI OD SEIZMICKIH SILA===================================================================

−Šemaopterećenja

−Klasifikacija konstrukcije prema osetljivostinahorizontalno pomeranje

H √ ∑Q

∑ Eb¿ I b≤0,2+0,1n (n≤3 )

≤0,6 (n≥4 )=¿Konstrukcija je sa nepomerljivimčvorovima

n−spratnostH−visinadeformabilnogdela konstrukcije

∑Q=(436,08+898,54+872,50+898,54+436,09 )+(97,46+212,69+205,96+212,69+97,49)

∑Q=4368,07 kN5∗Eb ¿ I b=5∗3,3∗107∗0,001251=206415 kN m2

Eb=3,3∗107 kN /m2 ( zaMB35 )

I b=0,354

12=0,001251m4

H=14 m

14∗√ 4368,07206415

=2,0366>0,6=¿Konstrukcija je sa pomerljivimčvorovima

STUB2−7−Izvijanje stubahi=h∗k

ψ=(∑ E∗I

h )s

(∑ E∗Il )

G

k=min(1+0,15∗(ψa+ψb)2+0,3∗ψ min )

I grede=0,3∗0,53

12=0,003125m4

I stuba=0,354

12=0,001251m4

ψ2=0 (uklještenje )

ψ7=( E∗I

h )2−7

+( E∗Ih )

7−12

(E∗Il )

6−7+( E∗I

l )7−8

=

0,0012515

+ 0,0012513

0,0031256

+ 0,0031256

=0,641

k=min(1+0,15∗(0+0,641 )=1,0962+0,3∗0=2 )=¿k=1,096

hi=1,096∗500cm=548,2 cm=5,482m

−vitkost stuba

λ=hi

imin= 5,482

0,1012=54,170=¿Oblast srednje vitkosti25<λ<75

imin=√ 0,0012510,352 =0,1012m

¿>Metodadopunske ekscentričnostie=e1+e0+eφ+eII

e1−ekscentricitet po teoriji I reda

e1=M q

Nq= 86,37

2153,87=0,0401m

e0−ekscentricitet usled netačnosti priizvođenju

e0=h

300= 5,0

300=0,01667m

eφ−dodatni ekscentricitet usled tečenjabetona1¿ λ≤50 (λ=54,170>50)=¿nije ispunjenuslov

2¿e1

d≥2( 0,0401

0,35=0,1146<2)=¿nije ispunjenus lov

3¿Ng≤0,2∗N q (898,54>0,2∗2153,87=430,774 )=¿ nije ispunjenuslov

eφ=(e1g+e0 )∗(2,718αE

1+αE∗φ

−1,0)e1g=

M g

N g= 1,50

898,54=0,0,0017

αE=N g

N E

N E=Eb¿ I b∗π2

hi2 =3,3∗107∗0,001251∗π2

5,4822 =13557,916 kN

αE=898,54

13557,916=0,0663

dm=2 Ab

O=2∗0,352

4∗0,35=0,175m

φ=3,6(iz tablice 11 ,PBAB 87)

eφ=(0,0017+0,0401 )∗(2,7180,0663

1+0,0663∗3,6−1,0)=0,0105m

eII−dodatni ekscentricitet II redae1

d=0,0401

0,35=0,1146 (0<

e1

d<0,3)

¿>eII=d∗λ−25

100∗√0,1+

e1

d=0,35∗54,17−25

100∗√0,1+ 0,0401

0,35=0,0473m

e=0,0401+0,01667+0,0105+0,0473=0,1146m e=0,1146m −DimenzionisanjeN u=2153,87kNM u=e∗N u=0,1146∗2153,87=246,834 kNm

mu=Mu

b∗d2∗f b= 246,834

0,35∗0,352∗23∗103 =0,250

nu=Nu

b∗d∗f b= 2153,87

0,35∗0,35∗23∗103 =0,764

¿>iz dijagramainterakcije μ=0,42 %

μ=μ∗f b

σ v=0,42∗23

240=0,0403

Aa2−7=μ∗b∗d=0,0403∗35∗35=49,31 cm2


ψ=(∑ E∗I

h )s

(∑ E∗Il )

G

k=min(1+0,15∗(ψa+ψb)2+0,3∗ψ min )

I grede=0,3∗0,53

12=0,003125m4

I stuba=0,354

12=0,001251m4

ψ7=( E∗I

h )2−7

+( E∗Ih )

7−12

(E∗Il )

6−7+( E∗I

l )7−8

=

0,0012515

+ 0,0012513

0,0031256

+ 0,0031256

=0,641

ψ12=( E∗I

h )7−12

+( E∗Ih )

12−17

( E∗Il )

11−12+(E∗I

l )12−13

=

0,0012513

+ 0,0012513

0,0031256

+ 0,0031256

=0.801

k=min(1+0,15∗(0.641+0,801 )=1.2162+0,3∗0.641=2.192 )=¿k=1.216

hi=1,216∗300cm=364.8 cm=3.648m

−vitkost stuba

λ=hi

imin= 3,648


imin=√ 0,0012510,352 =0,1012m



e1=M q

Nq= 54,06

1630,17=0,0332m

e0−ekscentricitet usled netačnosti priizvođenju

e0=h

300= 3,0

300=0,01m

eφ−dodatni ekscentricitet usled tečenjabetona1¿ λ≤50 (λ=36,05<50)=¿ ispunjenuslov

2¿e1

d≥2

3¿Ng≤0,2∗Nq

eφ=0m


d=0,0332

0,35=0,0949(0<

e1

d<0,3)

¿>eII=d∗λ−25

100∗√0,1+

e1

d=0,35∗36,05−25

100∗√0,1+ 0,0332

0,35=0,0171m

e=0,0332+0,01+0+0,0171=0,0603m e=0,0603m

−DimenzionisanjeNu=1630,17 kNM u=e∗N u=0,0603∗1630,17=98,299kNm

mu=Mu

b∗d2∗f b= 98,299

0,35∗0,352∗23∗103 =0,100

nu=Nu

b∗d∗f b= 1630,17

0,35∗0,35∗23∗103 =0,579

μ=μmin=1,000 %

Aa7−12= μ

100∗b∗d= 1

100∗35∗35=12,25 cm2


ψ=(∑ E∗I

h )s

(∑ E∗Il )

G

k=min(1+0,15∗(ψa+ψb)2+0,3∗ψ min )

I grede=0,3∗0,53

12=0,003125m4

I stuba=0,354

12=0,001251m4

ψ12=( E∗I

h )7−12

+( E∗Ih )

12−17

( E∗Il )

1−12+( E∗I

l )12−13

=

0,0012513

+ 0,0012513

0,0031256

+ 0,0031256

=0.801

ψ17=( E∗I

h )12−17

+( E∗Ih )

17−22

( E∗Il )

16−17+(E∗I

l )17−18

=

0,0012513

+ 0,0012513

0,0031256

+ 0,0031256

=0.801

k=min(1+0,15∗(0.801+0,801 )=1.2402+0,3∗0.801=2.240 )=¿k=1.240

hi=1,240∗300cm=372 cm=3.720m

−vitkost stuba

λ=hi

imin= 3,720


imin=√ 0,0012510,352 =0,1012m



e1=M q

Nq= 36,45

1127,85=0,0323m

e0−ekscentricitet usled netačnosti pri izvođenju

e0=h

300= 3,0

300=0,01m

eφ−dodatni ekscentricitet usled tečenjabetona1¿ λ≤50 (λ=36,76<50)=¿ ispunjenuslov

2¿e1

d≥2

3¿Ng≤0,2∗Nq

eφ=0m


d=0,0323

0,35=0,0923(0<

e1

d<0,3)

¿>eII=d∗λ−25

100∗√0,1+

e1

d=0,35∗36,76−25

100∗√0,1+ 0,0323

0,35=0,0180m

e=0,0323+0,01+0+0,0180=0,0603m e=0,0603m

−DimenzionisanjeN u=1127,85 kNM u=e∗N u=0,0603∗1127,85=68,01kNm

mu=Mu

b∗d2∗f b= 68,01

0,35∗0,352∗23∗103 =0,069

nu=Nu

b∗d∗f b= 1127,85

0,35∗0,35∗23∗103 =0,400

μ=μmin=1,000 %

Aa12−17= μ

100∗b∗d= 1

100∗35∗35=12,25 cm2


ψ=(∑ E∗I

h )s

(∑ E∗Il )

G

k=min(1+0,15∗(ψa+ψb)2+0,3∗ψ min )

I grede=0,3∗0,53

12=0,003125m4

I stuba=0,354

12=0,001251m4

ψ22=( E∗I

h )17−22

( E∗Il )

21−22+( E∗I

l )22−23

=

0,0012513

0,0031256

+ 0,0031256

=0.400

ψ17=( E∗I

h )12−17

+( E∗Ih )

17−22

( E∗Il )

16−17+(E∗I

l )17−18

=

0,0012513

+ 0,0012513

0,0031256

+ 0,0031256

=0.801

k=min(1+0,15∗(0,400+0,801 )=1,1802+0,3∗0.400=2.120 )=¿k=1,180

hi=1,180∗300cm=354 cm=3.540m

−vitkost stuba

λ=hiimin

= 3,5400,1012

=34,98=¿Oblast srednje vitkosti25<λ<75

imin=√ 0,0012510,352 =0,1012m



e1=M q

Nq= 30,30

630,52=0,0481m

e0−ekscentricitet usled netačnosti pri izvođenju

e0=h

300= 3,0

300=0,01m

eφ−dodatni ekscentricitet usled tečenjabetona

1¿ λ≤50 (λ=34,98<50)=¿ ispunjenuslov

2¿e1

d≥2

3¿Ng≤0,2∗Nq

eφ=0m


d=0,0481

0,35=0,137(0<

e1

d<0,3)

¿>eII=d∗λ−25

100∗√0,1+

e1

d=0,35∗34,98−25

100∗√0,1+ 0,0481

0,35=0,0170m

e=0,0481+0,01+0+0,0170=0,0751m e=0,0751m

−DimenzionisanjeN u=630,52kNM u=e∗N u=0,0751∗630,52=47,352kNm

mu=Mu

b∗d2∗f b= 47,352

0,35∗0,352∗23∗103 =0,048

nu=Nu

b∗d∗f b= 630,52

0,35∗0,35∗23∗103 =0,224

μ=μmin=1,000 %

Aa17−22= μ

100∗b∗d= 1

100∗35∗35=12,25 cm2

Aa=49,31 cm2

USVAJAM 8Φ32 (Aastv=64,34 cm2)

uzengije :Φu=Φmax

3=32

3=10.66=¿USVAJAMUΦ12/7 .5 /15cm

−Specifikacijaarmature

rednibroj Oblik i dimenzije ϕ

[mm]n

[kom ]ln[m]

Σln[m]

g¿kg/m]

G[kg]

1 32 6 2.66 15.96 6.313 100.76

2 32 6 7.91 47.46 6.313 299.61

3 32 6 9.20 55.20 6.313 348.48

4 32 6 6.95 41.58 6.313 262.49

5 32 6 5.90 35.40 6.313 223.48

6 32 6 2.90 17.40 6.313 109.85

7 12 155 1.76 272.80 0.888 242.25

ΣG 1586.92

1. Grafichkia Rad - AB STUB

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