1. graficki rad

[ ] 1. grafički rad

10

dp =

16

a/3=155 a/2=230

b0 x t0

h0 x ?0

b3 x t3b2 x t2b1 x t1

1:3

l=29 ma=4 .6 mφ=2.5

ε s=0.30MB=50

1. Elementi spregnutog preseka

1.1.Čelični nosač

Vertikalni lim ≠ h0 ∙ δ0

h0=( 115

÷1

25 ) ∙l= 120

∙l= 120

∙29=1.45 m→ usvojeno :h0=1450 mm

δ 0=8+ l750

=8+ 2900750

=11.87 m m→usvojeno : δ0=12 mm

Gornji pojas ≠ b0 ∙ t0

b0=min 250 mm →usvojeno :b0=300 mm

t 0=min 10 ÷ 12mm→ usvojeno : t0=14 mm

Donji pojas ≠ b1 ∙t 1

b1=h0

4+100 mm=1450

4+100=462.50 mm → usvojeno :b1=470 mm

t 1=min 20 mm →usvajam : t 1=25 mm

≠ b2 ∙t 2→ ≠ 420 x20

Marko Milošević MRG 52/09 1


≠ b3 ∙t 3 →≠ 370 x15

1.2.Armirano betonska ploča

d p=min 16 cm→ usvojeno :d p=16 cm

1.3.Moždanici

Prvo treba da izračunamo smičuće sile.

2. Merodavne veličine

2.1.Konstrukcioni čelik

Eč=210 GPa=2.1∙ 108 kN /cm2

νa=0.3

f a=7850 kg /m3

α t=1.2∙ 10−5

2.2.Beton

Eb=9.25 ∙ 3√MB+10=9.25∙ 3√50+10=36.21 GPa

νb=0.2

α t=1.2∙ 10−5

2.3.Efektivna širina betonskog preseka

bk bn bp

dph

v

155.00 230.00

be=bek+bn+bep≤ bstv

hv=10 cm

bk=a3−

b0

2−hv=

4603

−302

−10=128.33 cm≈ 130 cm

bn=b0+2 ∙ hv=30+2∙10=50 cm



b p=a2−

b0

2−hv=

4602

−302

−10=205 cm

βk=1

1+6.4 ∙( bk

l )2= 1

1+6.4 ∙( 1302900 )

2=0.987

β p=1

1+6.4 ∙( bp

l )2= 1

1+6.4 ∙( 2052900 )

2=0.969

bek=βk ∙bk=0.987 ∙ 130=128.31 cm

bep=β p ∙b p=0.969 ∙ 205=198.64 cm

be=128.31+50+198.64=376.95 cm

Za proračun uzimamo: bef=130+50+205=385 cm

3. Definisanje karakterističnih preseka

0

0

1

1

2

2

1

1

0

0

L/4=725

3L/8=1087.5

L/2=1450

presek "0" presek "1" presek "2"

?1450 x 12

?300 x 14

?470 x 25

?1450 x 12

?300 x 14

?470 x 25?420 x 20

?1450 x 12

?300 x 14

?470 x 25?420 x 20?370 x 15



4. Karakteristike preseka

4.1.Armirano betonska ploča

130.00 50.00 205.00

10

16

10

Y

X

16

30.00 30.00 185.0030.00110.00

385.00

y tež=385 ∙ 16 ∙ 8−110 ∙ 6

2∙ 14+2 ∙

30 ∙ 102

∙(16+ 103 )+30 ∙ 10 ∙(16+ 10

2 )385 ∙ 16−110 ∙ 6

2+2∙

30∙ 102

+30 ∙10=56760

6430=8.83cm

Ab=385 ∙16−110 ∙62

+2 ∙30 ∙10

2+30 ∙ 10=6430 cm2

I b=385 ∙163

12+385 ∙16 ∙ 0.832−110 ∙ 63

36−110 ∙ 6

2∙ 5.172+2 ∙( 30∙ 103

36+ 30 ∙ 10

2∙ 10.502)+ 30 ∙103

12+30 ∙10 ∙12.172=207 850.757 cm4

230

10 16

30 30 18530110

155

385

Tb

17.1

7 8.83

Za vreme t=0

uporednimodul elastičnosti : Ec=Ea→ νa=1.0

Ec=210GPa=2.1 ∙ 106 daN /cm2=2.1 ∙108 kN /cm2

Eb=9.25 ∙ 3√MB+10=9.25∙ 3√50+10=36.21 GPa=0.3621 ∙106daN /cm2=0.3621 ∙108kN /cm2

Koeficijent redukcije : νb=Eb

Ea

=0.3621∙106

2.1∙ 106 =0.172



Abr=νb ∙ Ab=0.172∙ 6430=1105.96 c m2

I br=νb ∙ I b=0.172 ∙207 850.757=35 750.33 cm4

Za vreme t=t n

φ=2.5

Ebφ=2

2+φ∙Eb=

22+2.5

∙ 36.21=16.09 GPa

νbφ=Ebφ

Ea

=16.09210

=0.077

Abrφ=νbφ ∙ Ab=0.077 ∙ 6430=495.11cm2

I brφ=νbφ ∙ I b=0.077 ∙207 850.757=16 004.51 cm4

Usvojen procenat armiranja : μ≈ 1 %→ Aa=1

100∙ Ab=

1100

∙ 6430=64.30 cm2

4.2.Čelični nosač

Presek „0”



Element Dimenzije A e A ∙ e e−e ' I a [cm4]

[mm] [mm] [cm2] [cm] [cm3] [cm] I as I ap

≠ b0 ∙ t0 300 14 45 17.87 750.54 89.97 6.86 339937.66

≠ h0 ∙ δ0 1450 12 174 91.07 15846.18 16.77 20.88 48907.05

≠ b1 ∙t 1 470 25 118 164.82 19366.35 56.98 61.20 381552.69∑ 334 107.84 35963.07 88.94 770397.40∑ 770486.34

Element A e A ∙ e e−e ' I a [cm4]

[cm2] [cm] [cm3] [cm] I as I ap

Armatura 64.30 0 0 90.43 0 525827.21Čelični nosač 334.00 107.84 35963.07 17.41 770486.34 101229.61

∑ 398.30 90.43 35963.07 770486.34 627056.81∑ 1397543.15

5. Idealni presek

Idealni presek, za vreme t=0

Ai 0=Aa+ Abr=398.30+1105.96=1504.26 cm2

yb=−∑ A ∙ c

Ai 0

=−35963.07

1504.26=−23.91 →T io

ya=c− yb=90.43−23.91=66.52 cm

I i 0=I a+ Aa ∙ ya2+ I br+ Abr ∙ yb

2

I i 0=1397543.15+398.30 ∙66.522+35750.33+1105.96 ∙ 23.912

I i 0=3827 999.46 cm4

Idealni presek, za vreme t=t n



Aiφ=Aa+ Abrφ=398.30+495.11=893.41 cm2

ηb=−∑ A ∙c

A iφ

=−35963.07

893.41=−40.25→ T iφ

ηa=c−ηb=90.43−40.25=50.18 cm

I iφ=I a+ Aa ∙ ηa2+ I brφ+Abrφ ∙ ηb

2

I iφ=1397543.15+398.30 ∙ 50.182+16004.51+495.11 ∙ 40.252

I iφ=3 218 589.11cm4

Presek „1”





≠ b0 ∙ t0 300 14 45 17.87 750.54 89.97 6.86 339937.66

≠ h0 ∙ δ0 1450 12 174 91.07 15846.18 16.77 20.88 48907.05

≠ b1 ∙t 1 470 25 118 164.82 19366.35 56.98 61.20 381552.69

≠ b2 ∙t 2 420 20 84 167.07 14033.88 59.23 28.00 294735.05∑ 418 119.75 49996.95 116.94 1065132.45∑ 1065249.39




∑ 481.80 103.77 49996.95 1065249.39 884796.07∑ 1950045.46


Ai 0=Aa+ Abr=481.80+1105.96=1587.76 cm2

yb=−∑ A ∙ c

Ai 0

=−49996.95

1587.76=−31.49 → T io

ya=c− yb=103.77−31.49=72.28 cm


2

I i 0=1950045.46+481.80∙72.282+35750.33+1105.96 ∙31.492



I i 0=5599 603.10 cm4


Aiφ=Aa+ Abrφ=481.80+495.11=976.91 cm2

ηb=−∑ A ∙c

A iφ

=−49996.95

976.91=−51.18 → T iφ

ηa=c−ηb=103.77−51.18=52.59 cm


2

I iφ=1950045.46+481.80 ∙52.592+16004.51+495.11 ∙51.182

I iφ=4595 455.50 cm4

Presek „2”





≠ b0 ∙ t0 300 14 45 17.87 750.54 89.97 6.86 339937.66

≠ h0 ∙ δ0 1450 12 174 91.07 15846.18 16.77 20.88 48907.05

≠ b1 ∙t 1 470 25 118 164.82 19366.35 56.98 61.20 381552.69

≠ b2 ∙t 2 420 20 84 167.07 14033.88 59.23 28.00 294735.05≠ b3 ∙t 3 370 15 56 168.82 9369.51 60.98 10.41 206411.97

∑ 473 125.51 59366.46 127.34 1271544.42∑ 1271671.76




∑ 537.30 110.49 59366.46 1271671.76 1107895.92∑ 2379567.69


Ai 0=Aa+ Abr=537.30+1105.96=1643.26 cm2



yb=−∑ A ∙ c

Ai 0

=−59366.46

1643.26=−36.13 →T io

ya=c− yb=110.49−36.13=74.36 cm


2

I i 0=2379567.69+537.30 ∙74.362+35750.33+1105.96 ∙ 36.132

I i 0=6829 964.43 cm4


Aiφ=Aa+ Abrφ=537.30+495.11=1 032.41 cm2

ηb=−∑ A ∙c

A iφ

=−59366.46

1032.41=−57.50 → T iφ

ηa=c−ηb=110.49−57.50=52.99 cm


2

I iφ=2379567.69+537.30 ∙ 52.992+16004.51+495.11 ∙ 57.502

I iφ=5 541 235.85 cm4

Presek „0“ „1“ „2“

1450 x 12

300 x 14

470 x 25

1450 x 12

300 x 14

470 x 25420 x 20

1450 x 12

300 x 14

470 x 25420 x 20370 x 15

Čeli

Nos Ač

[cm2]334 418 473



čni p

rese

k

ač

e '[cm ] 107.84 119.75 125.51

I č

[cm4]770 486.34 1 065 249.39 1 271 671.76

Nos

ač i

arm

atur

a Aa

[cm¿¿2]¿398.30 481.80 537.30

c[cm ] 90.43 103.77 110.49

I a

[cm4]1 397 543.15 1 950 045.46 2 379 567.69

Spre

gnuti

pre

sek

t=0

Ai 0

[cm¿¿2]¿1 504.26 1 587.76 1 643.26

− yb

[cm ]23.91 31.49 36.13

I i 0

[cm4]3 827 999.46 5 599 603.10 6 829 964.43

t=t n

Aiφ

[cm¿¿2]¿893.41 976.91 1 032.41

−ηb

[cm ]40.25 51.18 57.50

I iφ

[cm4]3 218 589.11 4 595 455.50 5 541 235.85

6. Proračun statičkih uticaja po fazama za statički sistem prosta greda

6.1.Proračun

Faza Ia – Montaža čeličnog nosača



RA RB

2900

Faza Ib – Betoniranje kolovozne ploče

Faza IIa – Spregnuti nosač prima uticaje sopstvene težineFaza IIb – Most opterećen pokretnim opterećenjemFaza IIIa – Spregnuti presek prima uticaje skupljanja betonaFaza IIIb – Spregnuti presek prima uticaje tečenja betonaFaza IIIc – Spregnuti presek prima uticaje različitog zagrevanja ploče i nosača

6.1.1. Faza I-a

Analiza opterećenja po jednom nosaču

Opis opterećenja A [cm2 ] γ [kN /m3 ] g1 [ kN /m' ]Čelični nosač 408.33 78.50 3.20

Kolovozna ploča 6430.00 25.00 16.08Oplata i skela 3.50

∑ q I=23.50 kN /m '

M q=2470.44 kNm



Ra=340.75 kN

6.1.2. Faza II-a

16

a/3=155 a/2=230

80.00 20.00 275.00

385.00

50.0

0

5.00

10.00


Opis opterećenja A [cm2 ] γ [kN /m3 ] g1 [ kN /m' ]Asfalt na kolovozu 1375.00 21.00 2.89Montažni AB venac 250.00 25.00 0.62

Beton ispune pešačke staze i instalacije 3.00Asfalt na pešačkoj stasi 160.00 21.00 0.34

Ograda 0.40Skela i oplata -3.50

∑ q I=3.75 kN /m '



M q=394.22 kNm

Ra=Rb=54.38 kN

6.1.3. Pokretno opterećenje



p =3 kN/m22

p =3 kN/m22

p =5 kN/m12 p =5 kN/m1

2

p =

50 k

N

p =

50 k

N

p =

50 k

N

1.50 1.50 1.50 1.50

2.00

3.00

6.00

p =50 kN p =50 kN p =50 kN

b 1b 1

p=3.00p=5.00

p=3.00

P=50.00

P=50.00

110.00 50.00 200.00 50.00 360.00

155.00 460.00 155.00

770.00

1.00

0.99

1.10

1.34 0.55

0.45

Ra

F1F2F3 -

0.34

Kd=1.4−0.008 ∙l=1.4−0.008 ∙29=1.168

P¿=P∙ (η1+η2 )∙ Kd=50 ∙ (0.99+0.55 ) ∙1.168=89.94 kN

p2¿=p2 ∙ ( F1+F2)=3 ∙( 1.34+1.10

2∙ 1.10+ 0.45

2∙2.05)=5.41 kN /m'

p1¿=p2

¿+ p1 ∙F3 ∙ Kd=5.41+5 ∙1.10+0.45

2∙3=17.04 kN /m'



p=17.04 p=17.04p=5.41

P=89.94

P=89.94

P=89.94

7.25

6.50

5.75

6.50

5.75

M (L/2)

F1 F2F3

11.50 1.50 1.50 1.50 1.50 11.50

29.00

M l /2p .o .=P¿ ∙ (η1+η2+η3 )+ p1

¿ ∙ ( F1+F2 )+ p2¿ ∙ F3=89.94 ∙ (7.25+6.50+6.50 )+17.04 ∙( 5.75

2∙11.50+ 5.75

2∙ 11.50)+5.41∙ 2 ∙

7.25+5.752

∙ 3=3159.04 kNm

p=17.04p=5.41

P=89.94

P=89.94

P=89.94

1.00

0.95

0.90

0.84

F1F2

Ta

1.50 1.50 1.50 24.50

29.00

T Ap .o .=P¿ ∙ (η1+η2+η3 )+ p1

¿ ∙F2+ p2¿ ∙ F1=89.94 ∙ (1.00+0.95+0.90 )+17.04 ∙

0.842

∙ 24.50+5.41∙1+0.84

2∙ 4.50=454.07 kN

p=17.04p=5.41

P=89.94

P=89.94

P=89.94

0.50

0.45

0.40

0.34

0.50

F1F2

T (l/2)

14.50 1.50 1.50 1.50 10.00

29.00



T l/2p .o .=P¿ ∙ (η1+η2+η3 )+ p1

¿ ∙F2+ p2¿ ∙ F1=89.94 ∙ (0.50+0.45+0.40 )+17.04 ∙

0.342

∙ 10+5.41∙0.50+0.34

2∙4.50=160.61 kN

6.2. Skupljanje betona

Ab=6430 cm2

O=1566.80 cm

dm=2 ∙ Ab

O= 2∙ 6430

1566.80=8.21

Relativna vlažnost sredine=90 %

ε s=0.30 ‰

t=t n

N sk=−Ebφ ∙ Ab ∙ εs=−1609 ∙ 6430 ∙0.0003=−3103.76 kN

M sk=N sk ∙ ηb=3103.76 ∙ 0.575=1784.662 kNm

6.3. Tečenje betona

ρ= φ2+φ

= 2.52+2.5

=0.56

M=M IIa=394.22 kNm

Nb 0=M IIa ∙Abr ∙ yb

Ii 0

=39422∙1105.96 ∙36.136829 964.43

=230.64 kN

Nφ=ρ ∙ Nb 0=0.56 ∙ 230.64=129.16 kN

M b 0=M b0 ∙I br

I i 0

=230.64 ∙35 750.33

6829 964.43=1.21 kNm

M φ=M +ρ ∙ ( M b 0+N b 0 ∙ ηb 0 )=394.22+0.56 ∙ (1.21+230.64 ∙0.5750 )=469.16 kNm

6.4. Temperaturna razlika

∆ t=10 °

ε t=αt ∙ ∆ t=1.2 ∙10−5 ∙ 10=1.2 ∙ 10−4

εt

εsk

=1.2∙10−4

1.5∙10−4 =0.80

N t=εt ∙ Ebφ ∙ Ab=1.2∙ 10−4 ∙1609 ∙6430=1241.50 kN

M t=N t ∙ ηb=± 1241.50 ∙0.5750=713.86 kNm



7. Proračun napona

7.1.Proračun napona u t=0

Faza I (nosač + ploča)

σ n=NAč

+ MIč

∙ y

N=0

M I=2470.44 kNm

σ nG= M

I č

∙ ynG= 247044

1271 671.76∙ (−108.34 )=−21.05 kN /cm2

σ nD=M

I č

∙ ynD= 247044

1271 671.76∙ 44.06=8.56 kN /cm2

Faza IIa (dodatno stalno opterećenje)

N=0

M IIa=394.22 kNm

σ bG=νb∙( N

Ai

+MI i

∙(− ybG))=0.172 ∙

394226829 964.43

∙ (−44.96 )=−0.045kN /cm2

σ bD=νb ∙( N

A i

+MIi

∙ (− ybG ))=0.172 ∙

394226829 964.43

∙ (−18.96 )=−0.019 kN /cm2

σ nG=νn∙( N

Ai

+MI i

∙ (− ybG))= 39422

6829964.43∙ (−18.96 )=−0.11kN /cm2

σ nD=νn ∙( N

A i

+MI i

∙ (+ ybG ))= 39422

6 829964.43∙ (+133.44 )=0.77kN /cm2

Faza IIb (pokretno opterećenje)

N=0

M IIb=3159.04 kNm

σ bG=νb ∙( N

Ai

+MI i

∙(− ybG ))=0.172 ∙

3159046 829 964.43

∙ (−44.96 )=−0.36 kN /cm2

σ bD=νb ∙( N

A i

+MIi

∙ (− ybG))=0.172 ∙

3159046829964.43

∙ (−18.96 )=−0.15kN /cm2



σ nG=νn ∙( N

Ai

+MI i

∙ (− ybG ))= 315904

6 829 964.43∙ (−18.96 )=−0.88 kN /cm2

σ nD=νn ∙( N

A i

+MI i

∙ (+ ybG))= 315904

6 829964.43∙ (+133.44 )=6.17kN /cm2



7.2.Proračun napona u t=t n

7.2.1. Uticaj skupljanja betona

M sk=1784.66 kNm N sk=−3103.76 kN

σ bG=νbφ ∙(N sk

A iφ

+M sk

I iφ

∙(−ηbG ))−(−ε sk )∙ Ebφ=0.077 ∙(−3103.76

1032.41+ 178466

5541235.85∙ (−66.33 ))− (−0.0003 ) ∙ 1609=−0.087 kN /cm2



σ bD=νbφ ∙( N sk

Aiφ

+M sk

Iiφ

∙ (−ηbD ))−(−εsk )∙ Ebφ=0.077 ∙(−3103.76

1 032.41+ 178466

5541235.85∙ (−40.33 ))−0.0003 ∙ 1609=−0.81 kN /cm2

σ nG=

N sk

A iφ

+M sk

I iφ

∙ (−ηnG )=−3103.76

1032.41+ 178466

5541 235.85∙ (−40.33 )=−4.30 kN /cm2

σ nD=

N sk

Aiφ

+M sk

I iφ

∙ ηnD =−3103.76

1032.41+ 178466

5541 235.85∙112.07=0.60 kN /cm2

7.2.2. Uticaj tečenja betona

M φ=469.16 kNm Nφ=−129.16 kN

σ bG=νbφ ∙( N φ

A iφ

+M φ

I iφ

∙ (−ηbG ))−ρ ∙ σb 0

G =0.077 ∙(−129.161032.41

+ 469.165 541235.85

∙ (−66.33 ))−0.56 ∙ 0.045=−0.035 kN /cm2

σ bG=νbφ ∙( N φ

A iφ

+M φ

I iφ

∙ (−ηbD ))− ρ∙ σ b0

D =0.077 ∙(−129.161032.41

+ 469.165 541 235.85

∙ (−40.33 ))−0.56 ∙0.019=−0.020 kN /cm2

σ nG=

N φ

A iφ

+M φ

I iφ

∙ ηnG =−129.16

1032.41+ 469.16

5 541235.85∙ (−40.33 )=−0.13kN /cm2

σ nD=

N sk

Aiφ

+M sk

I iφ

∙ ηnG =−129.16

1032.41+ 469.16

5 541 235.85∙ 112.07=−0.12 kN /cm2

7.2.3. Uticaj temperaturne razlike

εt

εsk

=1.2∙10−4

1.5∙10−4 =0.80

σ bG=σ b

sk ∙εt

εsk

=−0.87 ∙ 0.80=−0.70kN /cm2

σ bD=σb

sk ∙εt

ε sk

=−0.81∙ 0.80=−0.65kN /cm2

σ nG=σ n

sk ∙εt

εsk

=−4.30 ∙ 0.80=−3.44kN /cm2

σ nD=σn

sk ∙εt

ε sk

=0.60∙0.80=0.48kN /cm2


RA RBRC


8. Proračun spregnute konstrukcije sa privremenim osloncem u l /2

Faza Ia – Montaža čeličnog nosača


RA RBRC

R'A R'BRC

R'A R'B

RC


l /2=14.5 m l /2=14.5 m

Faza Ib – Betoniranje kolovozne ploče

l /2=14.5 m l /2=14.5 m

Faza IIa – Po očvršćavanju betona uklanja se privremeni oslonac a njegovu reakciju prima spregnuti nosač

l /2=14.5 m l /2=14.5 m

Faza IIb – Spregnuti nosač prima uticaje sopstvene težine druge faze (asfalt, ispuna, pešačke staze, AB venac, ograda)Faza IIc – Most je opterećen pokretnim opterećenjemFaza IIIa – Spregnuti presek prima uticaje skupljanja betonaFaza IIIb – Spregnuti presek prima uticaje tečenja betona



Faza IIIc – Spregnuti presek prima uticaje različitog zagrevanja kolovozne ploče i nosača

8.1.Statički uticaji po fazama

8.1.1. Faza I




∑ q I=23.50 kN /m 'p=23.50

14.50 14.50

347.39

-617.61

347.39

-127.79

212.96

-212.96

127.79

M C=−q ∙( l

2 )2

8=−23.5 ∙14.52

8=−617.61 kNm

M max=347.39 kNm

RA=RB=0.375 ∙ q ∙l2=0.375 ∙ 23.50 ∙14.5=127.78 kN

RC=1.25∙ q ∙l2=1.25 ∙ 23.5 ∙14.5=425.94 kN

8.1.2. Faza IIa



P=425.94

29.00

3088.06

-212.97

212.97

M max=RC ∙ l

4= 425.94 ∙29

4=3 088.06 kNm

RA=RB=RC

2=425.94

2=212.97 kN

8.1.3. Faza IIb

9. Analiza opterećenja po jednom nosaču




∑ q II=3.75 kN /m '



8.1.4. Faza IIc – Pokretno opterećenje

p =3 kN/m22

p =3 kN/m22

p =5 kN/m12 p =5 kN/m1

2

p =

50 k

N

p =

50 k

N

p =

50 k

N

1.50 1.50 1.50 1.50

2.00

3.00

6.00

p =50 kN p =50 kN p =50 kN

b 1b 1

p=3.00p=5.00

p=3.00

P=50.00

P=50.00

110.00 50.00 200.00 50.00 360.00

155.00 460.00 155.00

770.00

1.00

0.99

1.10

1.34 0.55

0.45

Ra

F1F2F3 -

0.34



Kd=1.4−0.008 ∙l=1.4−0.008 ∙29=1.168

P¿=P∙ (η1+η2 )∙ Kd=50 ∙ (0.99+0.55 ) ∙1.168=89.94 kN

p2¿=p2 ∙ ( F1+F2)=3 ∙( 1.34+1.10

2∙ 1.10+ 0.45

2∙2.05)=5.41 kN /m'

p1¿=p2

¿+ p1 ∙F3 ∙ Kd=5.41+5 ∙1.10+0.45

2∙3=17.04 kN /m'

p=17.04 p=17.04p=5.41

P=89.94

P=89.94

P=89.94

7.25

6.50

5.75

6.50

5.75

M (L/2)

F1 F2F3

11.50 1.50 1.50 1.50 1.50 11.50

29.00

M l /2p .o .=P¿ ∙ (η1+η2+η3 )+ p1

¿ ∙ ( F1+F2 )+ p2¿ ∙ F3=89.94 ∙ (7.25+6.50+6.50 )+17.04 ∙( 5.75

2∙11.50+ 5.75

2∙ 11.50)+5.41∙ 2 ∙

7.25+5.752

∙ 3=3159.04 kNm

p=17.04p=5.41

P=89.94

P=89.94

P=89.94

1.00

0.95

0.90

0.84

F1F2

Ta

1.50 1.50 1.50 24.50

29.00

T Ap .o .=P¿ ∙ (η1+η2+η3 )+ p1

¿ ∙F2+ p2¿ ∙ F1=89.94 ∙ (1.00+0.95+0.90 )+17.04 ∙

0.842

∙ 24.50+5.41∙1+0.84

2∙ 4.50=454.07 kN

8.2. Faza IIIa – Skupljanje betona

Ab=6430 cm2

O=1566.80 cm

dm=2 ∙ Ab

O= 2∙ 6430

1566.80=8.21




ε s=0.30 ‰

t=t n

N sk=−Ebφ ∙ Ab ∙ εs=−1609 ∙ 6430 ∙0.0003=−3103.76 kN

M sk=N sk ∙ ηb=3103.76 ∙ 0.575=1784.662 kNm

8.3. Faza IIIb – Tečenje betona

ρ= φ2+φ

= 2.52+2.5

=0.56

M=M IIa+M IIb=3088.06+394.22=3482.28 kNm

Nb 0=M ∙Abr ∙ ybr

I i0

=348228 ∙1105.96 ∙36.13

6 829 964.43=2037.29 kN

Nφ=ρ ∙ Nb 0=0.56 ∙ 2037.29=1140.88kN

M b 0=M ∙I br

I i 0

=3482.28 ∙35 750.33

6 829 964.43=18.23 kNm

M φ=M +ρ ∙ ( M b 0+N b 0 ∙ ηb 0 )=3482.28+0.56 ∙ (18.23+2037.29 ∙0.5750 )=4148.50 kNm

8.4. Faza IIIc – Temperaturna razlika

∆ t=10 °

ε t=αt ∙ ∆ t=1.2 ∙10−5 ∙ 10=1.2 ∙ 10−4

εt

εsk

=1.2∙10−4

1.5∙10−4 =0.80

N t=εt ∙ Ebφ ∙ Ab=1.2∙ 10−4 ∙1609 ∙6430=1241.50 kN

M t=N t ∙ ηb=± 1241.50 ∙0.5750=713.86 kNm

9. Proračun napona u l /2

9.1. Proračun napona u t=09.1.1. Faza I – Uticaje prima čelični nosač

M CI =−617.61 kNm

N I=0



σ nG=

M I

I io

∙ ynG= −61761

1271671.76∙ (−108.34 )=5.26 kN /cm2

σ nD=

M I

I io

∙ ynD= −61761

1271671.76∙ 44.06=−2.14 kN /cm2

9.1.2. Faza IIa + IIb – Uticaje prima spregnuti nosač

M II=M IIa+ M IIb=3088.06+394.22=3482.28 kNm

N II=0

σ bG=νb

M I

I io

∙ ynG=0.172 ∙

3482286829964.43

∙ (−44.96 )=−0.39 kN /cm2

σ bD=νb

M I

I io

∙ ynD=0.172 ∙

3482286829964.43

∙ (−18.96 )=−0.17kN /cm2

σ nG=

M II

I io

∙ ynG= 348228

6829964.43∙ (−18.96 )=−0.97 kN /cm2

σ nD=

M II

I io

∙ ynD= 348228

6829964.43∙ 133.44=6.80 kN /cm2


M=3159.04 kNm

N=0

σ bG=νb

M I

I io

∙ ynG=0.172 ∙

3159046829964.43

∙ (−44.96 )=−0.36 kN /cm2

σ bD=νb

M I

I io

∙ ynD=0.172 ∙

3159046829964.43

∙ (−18.96 )=−0.15kN /cm2

σ nG=

M II

I io

∙ ynG= 315904

6829964.43∙ (−18.96 )=−0.88 kN /cm2

σ nD=

M II

I io

∙ ynD= 315904

6829964.43∙ 133.44=6.17 k N /cm2



9.2.Proračun napona u t=t n

9.2.1. Faza IIIa – Skupljanje betona

M sk=1784.66 kNm

N sk=−3103.76 kN

σ bG=νbφ( N sk

A iφ

+M sk

I iφ

∙ ηG)−εsk ∙ Ebφ=0.077 ∙(−3103.761032.41

+ 1784665541235.85

∙ (−66.33 ))−0.0003 ∙ 1609=−0.88 kN /cm2

σ bD=νbφ (N sk

A iφ

+M sk

I iφ

∙ ηD)−εsk ∙ Ebφ=0.077 ∙(−3103.761032.41

+ 1784665541235.85

∙ (−40.33 ))−0.0003 ∙1609=−0.81 kN /cm2

σ nG=

N sk

A iφ

+M sk

I iφ

∙ ηnG=−3103.76

1032.41+ 178466

5541235.85∙ (−40.33 )=−4.30 kN /cm2

σ nD=

N sk

Aiφ

+M sk

I iφ

∙ ηnD=−3103.76

1032.41+ 178466

5541235.85∙112.07=0.60 kN /cm2

9.2.2. Uticaj tečenja

M φ=4148.50 kNm

Nφ=1140.88 kN

σ bG=νbφ( Nφ

Aiφ

+M φ

I iφ

∙ ηG)− ρ∙ σb 0=0.077 ∙( 1140.881032.41

+ 4148505541235.85

∙ (−66.33 ))−0.56 ∙ 0.39=−0.51kN /cm2

σ bD=νbφ ( N φ

A iφ

+M φ

I iφ

∙ ηD)−ρ∙ σ b 0=0.077 ∙( 1140.881032.41

+ 4148505541235.85

∙ (−40.33 ))−0.56 ∙ 0.17=−0.24 kN /cm2

σ nG=

N φ

A iφ

+M φ

I iφ

∙ ηG= 1140.881032.41

+ 4148505541235.85

∙ (−40.33 )=−1.91 kN /cm2

σ nD=

Nφ

Aiφ

+M φ

I iφ

∙ ηD= 1140.881032.41

+ 4148505541235.85

∙112.07=9.49 kN /cm2



9.2.3. Faza IIIc – Uticaj temperaturne razlike

εt

εsk

=1.2∙10−4

1.5∙10−4 =0.80

σ bG=σ b

sk ∙εt

εsk

=−0.87 ∙ 0.80=−0.70kN /cm2

σ bD=σb

sk ∙εt

ε sk

=−0.81∙ 0.80=−0.65kN /cm2

σ nG=σ n

sk ∙εt

εsk

=−4.30 ∙ 0.80=−3.44kN /cm2

σ nD=σn

sk ∙εt

ε sk

=0.60∙0.80=0.48kN /cm2

FazeNapon I esk g + tečenje p Osnovno

opterećenje ΔtOsnovno

+dopunsko

σ bG -0.88 -0.51 -0.36 -1.75 -0.70 -2.45

σ bD -0.81 -0.24 -0.15 -1.20 -0.65 -1.85

σ nG 5.26 -4.30 -1.91 -0.88 -1.83 -3.44 -5.27

σ nD -2.14 0.60 9.49 6.17 14.12 0.48 14.60

Za čelik Č.0461 → σv=26 kN /cm2

Osnovno opterećenje: σ dop=0,665 ∙ σv=0,665 ∙ 26=17,29 kN /cm2

Totalno opterećenje: σ dop=0,75 ∙ σv=0,75 ∙ 26=19,50 kN /cm2

MB 50: σ dop=¿

σ dop=17.29 kN /cm2>14.12 kN /cm2

σ dop=19.50 kN /cm2>14.60 kN /cm2



10. Proračun spregnute konstrukcije sa podizanjem u l /2

Faza Ia – Montaža čeličog nosača preko privremenog oslonca u l /2

RA RBRC

Faza Ib – Nadvišenje čeličnog nosača dejstvom hudrauličnih presa u l /2

RA RBRC+P

Faza Ic – Betoniranje kolovozne ploče

RA RBRC+P'' ''



Faza IIa – Posle vezivanja betona i uklanjanja oslonca, spregnuti presek prima reakciju Rc+P

R'A R'B

RC+P

Faza IIb – Spregnuti presek prima uticaje sopstvene težine II fazeFaza IIc – Most opterećen pokretnim opterećenjemFaza IIIa – Spregnuti presek prema uticajima skupljanja betonaFaza IIIb – Spregnuti presek prema uticajima tečenja betonaFaza IIIc – Spregnuti presek prema uticajima različitog zagrevanja kolovoza i čeličnog nosača

10.1. Faza I10.1.1. Faza Ia




∑ q I=23.50 kN /m '



p=23.50

14.50 14.50

347.39

-617.61

347.39

-127.79

212.96

-212.96

127.79

M C=−q ∙( l

2 )2

8=−23.5 ∙14.52

8=−617.61 kNm

M max=347.39 kNm

RA=RB=0.375 ∙ q ∙l2=0.375 ∙ 23.50 ∙14.5=127.78 kN

RC=1.25∙ q ∙l2=1.25 ∙ 23.5 ∙14.5=425.94 kN

10.1.2. Faza Ib – Apliciranje potrebne sile u presi iz uslova maksimalnog iskorišćenja napona čelika

σ č=σ doz=MI č

∙ y → M max=σdoz ∙ I č

yčG =17.29∙ 1271671.76

108.34=202946.32 kNcm=2029.46 kNm

−q ∙ l2

8−

Pmax ∙ l¿

4≤ M max=−2029.46

−23.5 ∙14.52

8−

Pmax ∙29

4=−2029.46

−Pmax ∙29

4=−2029.46+617.61

Pmax=194.74 kN



Usvajam: P=50 kN

M I=q ∙l2

8+ P ∙l¿

4=23.5 ∙ 14.52

8+ 50 ∙29

4=980.11kNm

P=50.00

14.50 14.50

-362.50

10.2. Faza II10.2.1. Faza IIa

Rc+P=425.94+50=475.94 kNP=475.94

14.50 14.50

3450.56

M IIa=3450.56 kNm

10.2.2. Faza IIb – Dodatna sopstvena težina





∑ q II=3.75 kN /m '



M IIb=394.22 kNm


p =3 kN/m22

p =3 kN/m22

p =5 kN/m12 p =5 kN/m1

2

p =

50 k

N

p =

50 k

N

p =

50 k

N

1.50 1.50 1.50 1.50

2.00

3.00

6.00

p =50 kN p =50 kN p =50 kN

b 1b 1



p=3.00p=5.00

p=3.00

P=50.00

P=50.00

110.00 50.00 200.00 50.00 360.00

155.00 460.00 155.00

770.00

1.00

0.99

1.10

1.34 0.55

0.45

Ra

F1F2F3 -

0.34

Kd=1.4−0.008 ∙l=1.4−0.008 ∙29=1.168

P¿=P∙ (η1+η2 )∙ Kd=50 ∙ (0.99+0.55 ) ∙1.168=89.94 kN

p2¿=p2 ∙ ( F1+F2)=3 ∙( 1.34+1.10

2∙ 1.10+ 0.45

2∙2.05)=5.41 kN /m'

p1¿=p2

¿+ p1 ∙F3 ∙ Kd=5.41+5 ∙1.10+0.45

2∙3=17.04 kN /m'

p=17.04 p=17.04p=5.41

P=89.94

P=89.94

P=89.94

7.25

6.50

5.75

6.50

5.75

M (L/2)

F1 F2F3

11.50 1.50 1.50 1.50 1.50 11.50

29.00

M l /2p .o .=P¿ ∙ (η1+η2+η3 )+ p1

¿ ∙ ( F1+F2 )+ p2¿ ∙ F3=89.94 ∙ (7.25+6.50+6.50 )+17.04 ∙( 5.75

2∙11.50+ 5.75

2∙ 11.50)+5.41∙ 2 ∙

7.25+5.752

∙ 3=3159.04 kNm



p=17.04p=5.41

P=89.94

P=89.94

P=89.94

1.00

0.95

0.90

0.84

F1F2

Ta

1.50 1.50 1.50 24.50

29.00

T Ap .o .=P¿ ∙ (η1+η2+η3 )+ p1

¿ ∙F2+ p2¿ ∙ F1=89.94 ∙ (1.00+0.95+0.90 )+17.04 ∙

0.842

∙ 24.50+5.41∙1+0.84

2∙ 4.50=454.07 kN

10.3.1. Faza IIIa – Uticaj skupljanja betona

Ab=6430 cm2

O=1566.80 cm

dm=2 ∙ Ab

O= 2∙ 6430

1566.80=8.21


ε s=0.30 ‰

t=t n

N sk=−Ebφ ∙ Ab ∙ εs=−1609 ∙ 6430 ∙0.0003=−3103.76 kN

M sk=N sk ∙ ηb=3103.76 ∙ 0.575=1784.662 kNm

10.3.2. Faza IIIb - Uticaj tečenja betona

ρ= φ2+φ

= 2.52+2.5

=0.56

M=M IIa+M IIb=3450.56+349.22=3799.78 kNm

Nb 0=M ∙Abr ∙ yb

I i0

=379978 ∙1105.96 ∙ 36.13

6829964.43=2223.04 kN

Nφ=ρ ∙ Nb 0=0.56 ∙ 2223.04=1111.52kN

M b 0=M ∙I br

I i 0

=3799.78 ∙35750.33

6829964.43=19.89 kNm



M φ=M +ρ ∙ ( M b 0+N b 0 ∙ ηb )=3799.78+0.56 ∙ (19.89+2223.04 ∙ 0.5750 )=4526.74 kNm


∆ t=10 °

ε t=αt ∙ ∆ t=1.2 ∙10−5 ∙ 10=1.2 ∙ 10−4

εt

εsk

=1.2∙10−4

1.5∙10−4 =0.80

N t=εt ∙ Ebφ ∙ Ab=1.2∙ 10−4 ∙1609 ∙6430=1241.50 kN

M t=N t ∙ ηb=± 1241.50 ∙0.5750=713.86 kNm

11. Proračun napona u l /2

11.1. Proračun napona u t=011.1.1. Faza I – Uticaje prima čelični nosač

M I=−q ∙( l

2 )2

8−P ∙l

4=−23.5 ∙14.52

8−50 ∙29

4=−617.61−362.5=−980.11kNm

σ nG= M

I č

∙ yG= −980111271671.76

∙ (−108.34 )=8.35kN /cm2

σ nD=M

I č

∙ y D=−98011

1271671.76∙ 44.06=−3.40 kN /cm2

11.1.2. Faza IIa + IIb

M=M IIa+M IIb=3450.56+394.22=3844.78 kNm

σ bG=νb

M I

I io

∙ ybG=0.172 ∙

3844786829964.43

∙ (−44.96 )=−0.44 kN /cm2

σ bD=νb

M I

I io

∙ ybD=0.172 ∙

3844786829964.43

∙ (−18.96 )=−0.18kN /cm2

σ nG=

M II

I io

∙ ynG= 384478

6829964.43∙ (−18.96 )=−1.07 kN /cm2

σ nD=

M II

I io

∙ ynD= 384478

6829964.43∙ 133.44=7.51 kN /cm2




M=3159.04 kNm

N=0

σ bG=νb

M I

I io

∙ ynG=0.172 ∙

3159046829964.43

∙ (−44.96 )=−0.36 kN /cm2

σ bD=νb

M I

I io

∙ ynD=0.172 ∙

3159046829964.43

∙ (−18.96 )=−0.15kN /cm2

σ nG=

M II

I io

∙ ynG= 315904

6829964.43∙ (−18.96 )=−0.88 kN /cm2

σ nD=

M II

I io

∙ ynD= 315904

6829964.43∙ 133.44=6.17 kN /cm2

11.2. Proračun napona u t=t n

11.2.1. Faza IIIa – Skupljanje betona

M sk=1784.66 kNm

N sk=−3103.76 kN

σ bG=νbφ( N sk

A iφ

+M sk

I iφ

∙ ηG)−εsk ∙ Ebφ=0.077 ∙(−3103.761032.41

+ 1784665541235.85

∙ (−66.33 ))−0.0003 ∙ 1609=−0.88 kN /cm2

σ bD=νbφ (N sk

A iφ

+M sk

I iφ

∙ ηD)−εsk ∙ Ebφ=0.077 ∙(−3103.761032.41

+ 1784665541235.85

∙ (−40.33 ))−0.0003 ∙1609=−0.81 kN /cm2

σ nG=

N sk

A iφ

+M sk

I iφ

∙ ηnG=−3103.76

1032.41+ 178466

5541235.85∙ (−40.33 )=−4.30 kN /cm2

σ nD=

N sk

Aiφ

+M sk

I iφ

∙ ηnD=−3103.76

1032.41+ 178466

5541235.85∙112.07=0.60 kN /cm2

11.2.2. Sopstvena težina + tečenje

M φ=4526.74 kNm

Nφ=1111.52kN

σ bG=νbφ( Nφ

Aiφ

+M φ

I iφ

∙ ηG)− ρ∙ σb 0=0.077 ∙( 1111.521032.41

+ 4526745541235.85

∙ (−66.33 ))−0.56 ∙ 0.44=−0.58 kN /cm2

σ bD=νbφ ( N φ

A iφ

+M φ

I iφ

∙ ηD)−ρ∙ σ b 0=0.077 ∙( 1111.521032.41

+ 4526745541235.85

∙ (−40.33 ))−0.56 ∙ 0.18=−0.27 kN /cm2



σ nG=

N φ

A iφ

+M φ

I iφ

∙ ηG= 1111.521032.41

+ 4526745541235.85

∙ (−40.33 )=−2.22 kN /cm2

σ nD=

Nφ

Aiφ

+M φ

I iφ

∙ ηD= 1111.521032.41

+ 4526745541235.85

∙112.07=10.23 kN /cm2


εt

εsk

=1.2∙10−4

1.5∙10−4 =0.80

σ bG=σ b

sk ∙εt

εsk

=−0.87 ∙ 0.80=−0.70kN /cm2

σ bD=σb

sk ∙εt

ε sk

=−0.81∙ 0.80=−0.65kN /cm2

σ nG=σ n

sk ∙εt

εsk

=−4.30 ∙ 0.80=−3.44kN /cm2

σ nD=σn

sk ∙εt

ε sk

=0.60∙0.80=0.48kN /cm2



12. Proračun spregnute konstrukcije raspona l+l (greda na dva polja)

Faza Ia – Postavljanje čeličnog nosačaFaza Ib – Betoniranje kolovozne pločeFaza Ic – Dodatno stalno opterećenjeFaza II – Pokretno opterećenjeFaza IIIa – Uticaj skupljanja betonaFaza IIIb – Uticaj tečenja betonaFaza IIIc – Uticaj temperaturne razlike

12.1. Faza I - Uticaj sopstvene težine




∑ q I=23.50 kN /m 'p=23.50

29.00 29.00

1389.48

-2470.34

1389.48

-255.57

425.93

-425.93

255.57

M C=−0.125 ∙ q ∙ l2=−0.125 ∙ 23.50∙292=−2470.34 kNm

RA=RB=0.375 ∙ q ∙l=0.375 ∙23.50 ∙ 29=255.57 kN

RC=1.25∙ q ∙ l=1.25 ∙ 23.50∙ 29=851.86 kN

M max=0.07 ∙ q ∙l2=0.07 ∙23.50 ∙ 292=1389.48 kNm


p*1 p*1p*2

P*

P*

P*

10

2 3 4 5 6 7 8 910

11 12 13 14 15 16 17 18 1920

29.00 29.00

F1F2 F3 F4


12.2. Faza II - Uticaj pokretnog opterećenja

η ' η1 η2

η ' '

2.90 :0.0022=1.40 : x → x=0.0011→η1=0.0938+0.0011=0.0949

2.90 :0.0022=0.10: x→ x=0.0001→ η '=0.0938−0.0001=0.0937

2.90 :0.0127=1.40 : x→ x=0.0061 →η2=0.0833+0.0061=0.0894

2.90 :0.0127=0.10 : x→ x=0.0004 → η ' '=0.0833−0.0004=0.0829


2.90

0.00

22

1.400.1

0.0960

0.0938

0.0960

0.0833

0.01

27

1.40 0.10

2.90

0 0 01 0.

0248

0.7192

2 0.0480

1.3920

3 0.0683

1.9807

4 0.0840

2.4360

η ' 0.0937

2.7173

5 0.0938

2.7202

η1 0.0949

2.7521

6 0.0960

2.7840

η2 0.0894

2.5926

7 0.0833

2.4157

η ' '0.0829

2.4041

8 0.0720

2.0880

9 0.0428

1.2412

10

0 0

11

0.0428

1.2412

12

0.0720

2.0880

13

0.0833

2.4157

14

0.0960

2.7840

15

0.0938

2.7202

16

0.0840

2.4360

17

0.0683

1.9807

18

0.0480

1.3920

19

0.02

0.71

F1

0.27193.83064.15214.03243.50580.2410

16.0338

F2

1.04283.06124.89046.40427.2146

22.6132

F3

6.28894.82731.7997

12.9159

F4

1.79974.82736.53047.53967.98117.47656.40424.89043.06121.0428

51.5532

p*1 p*2

P*

P*

P*

10 2 3 4 5 6 7 8 9 1011 12 13 14 15 16 17 18 19

20

29.00 29.00

F1

F2 F3

p*1


10

2 3 4 5 6 7 8 910

11 12 13 14 15 16 17 18 1920F1F2 F3 F4

M 10=−(P¿( η1+η2+η3 )+ p1¿ ( F2+F3+F4 )+ p2

¿ ∙ F1)=−( 89.94 ∙ (2.7521+2.7840+2.5926 )+17.04 ∙ (22.6132+12.9159+51.5532 )+5.41 ∙16.0338 )=−2301.72 kNm

M 10=−2301.72 kNm

η ' η1 η2 η ' '

2.90

0.2064

0.1527

1.400.10

0.05

37

2.90

0.2064

0.1625

1.40 0.100.

0439

2.90 :0.0537=1.40 : x→ x=0.0259 → η1=0.1527+0.0259=0.1786

2.90 :0.0537=0.10 : x→ x=0.0018 → η '=0.1527−0.0018=0.1509

2.90 :0.0439=1.40: x→ x=0.0212→ η2=0.1625+0.0212=0.1837

2.90 :0.0439=0.10 : x→ x=0.0015 → η' '=0.1625−0.0212=0.1413


0 0 01 0.

0501

1.4529

2 0.1008

2.9232

η ' 0.1509

4.3761

3 0.1527

4.4283

η1 0.1786

5.1794

4 0.2064

5.9856

η2 0.1837

5.3273

5 0.1625

4.7125

η ' '0.1413

4.0977

6 0.1216

3.5264

7 0.0843

2.4447

8 0.0512

1.4848

9 0.0229

0.6641

10

0 0

F1

0.44026.72548.37388.48477.02790.4405

31.4925

F2

2.10676.3453

10.219018.6710

F3

10.67378.65815.69783.11590.9629

29.1087

p*1p*2

P*

P*

P*

10 2 3 4 5 6 7 8 9 1011 12 13 14 15 16 17 18 19

20

29.00 29.00

F1F2


10 2 3 4 5 6 7 8 9 1011 12 13 14 15 16 17 18 19

20

F1

F2 F3

M 10=−(P¿( η1+η2+η3 )+ p1¿ ( F2+F3 )+ p2

¿ ∙F1 )=−( 89.94 ∙ (5.1794+5.9856+5.3273 )+17.04 ∙ (18.6710+29.1087 )+5.41∙ 31.4925 )=−2467.86 kNm

Q0

η ' η1 η2 η ' '

2.90

1.40

1.00

0.8753

0.1

24

7

2.90

0.8753

0.7520

1.500.10

0.1

23

3

2.90 :0.1247=1.40 : x→ x=0.0602 →η1=0.8753+0.0602=0.9355

2.90 :0.1233=2.80 : x→ x=0.1190→ η2=0.7520+0.1190=0.8710

2.90 :0.1233=1.30 : x→ x=0.0553 →η ' '=0.7520+0.0553=0.8073


0 1η1 0.935

51 0.875

3η2 0.871

0η ' ' 0.807

32 0.752

03 0.613

24 0.516

05 0.406

06 0.304

07 0.211

08 0.128

09 0.057

010 0

F1

1.45161.26760.08731.25874.0652

F2

1.01351.97951.63731.33691.02950.74240.49160.26820.08268.5815

p*1p*2

P*

P*

P*

10 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 1920

29.00 29.00

p*1


10 2 3 4 5 6 7 8 9 1011 12 13 14 15 16 17 18 19

20

F1F2

Q0=P¿ ∙ (η '+η1+η2 )+ p1¿ ∙F2+ p2

¿ ∙ F3=89.94 ∙ (1.0+0.9355+0.8710 )+17.04 ∙8.5815+5.41∙ 4.0652=420.6381 kN

Q10

η ' η1 η2 η ' '

1.00

2.90

1.40

0.943

0.05

7

0.943

2.90

0.872

1.500.10

0.07

1

2.90 :0.057=1.40 : x→ x=0.028 → η1=0.943+0.028=0.971

2.90 :0.071=2.80: x → x=0.068 →η2=0.872+0.068=0.940


0 01 0.0252 0.0483 0.0684 0.0845 0.0946 0.0967 0.0898 0.0729 0.043

10 1η1 0.971

11 0.943η2 0.940

η ' ' 0.90412 0.87213 0.78914 0.69615 0.59416 0.48417 0.36818 0.24819 0.12520 0

F1

1.4781.3400.0941.3834.295

F2

1.1542.4081.4111.8751.5631.2350.8930.5410.181

11.261

F3

0.0360.1060.1680.2200.2580.2760.2680.2330.1670.0621.794


10 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 1920

F1F2

Q10=P¿ ∙ (η '+η1+η2 )+ p1¿ ∙ ( F2+F3 )+ p2

¿ ∙ F1=89.94 ∙ (1.0+0.971+0.940 )+17.04 ∙ (11.261+1.794 )+5.41 ∙ 4.295=507.508 kN

12.3. Uticaj skupljanja betona

N sk=−Ebφ ∙ Ab ∙ εs=−16.09 ∙106 ∙0.643 ∙0.30 ∙10−3=−3103.76 kN

M sk=N sk ∙ ηb=−3103.76 ∙ (−0.575 )=1784.66 kNm


Tb

Ti?

Nsk

Msk 57.

50

Msk=Mo

2929

M 1

M

892.79

1.00

1784.66 1784.66

X 1=1.0


X1φ=−δ 10

δ 11

=−δ 1 sk

δ11 φ

E I c δ 11φ=∫M 2 I c

Ids=1

3∙l ∙1.0 ∙ 2=1

3∙29 ∙1 ∙ 2=19.33

E I c δ 1 sk=∫ M M sk

I c

Ids=1

2∙ l ∙ 1.0 ∙ M sk ∙2=1

2∙ 29 ∙1 ∙ 1784.66 ∙2=51755.14

X1φ=−51755.14

19.33=−2677.45

M=M sk+M ∙ X 1φ=1784.66−1 ∙2677.45=−892.79 kNm

12.4. Faza IIIc - Uticaj tečenja betona

ρ= φ2+φ

= 2.52+2.5

=0.56

M=M g=2470.34 kNm

Nb 0=M ∙Abr ∙ yb

I i0

=247034 ∙1105.96 ∙36.13

6829964.43=1445.26 kN

Nφ=ρ ∙ Nb 0=0.56 ∙ 1445.26=809.36 kN

M b 0=M ∙I br

I i 0

=2470.34 ∙35750.33

6829964.43=12.93 kNm

M φ=M +ρ ∙ ( M b 0+N b 0 ∙ ηb )=2470.34+0.56 ∙ (12.93+1445.26 ∙ 0.5750 )=2942.95 kNm

13. Proračun napona za l+l

13.1. Proračun u t=013.1.1. Faza I – Uticaj sopstvene težine

M g=−2470.34 kNm

σ bG=νb

M g

I io

∙ ybG=0.172∙

−2470346829964.43

∙ (−44.96 )=0.28 kN /cm2

σ bD=νb

M g

I io

∙ ybD=0.172 ∙

−2470346829964.43

∙ (−18.96 )=0.12kN /cm2

σ nG=

M g

Iio

∙ ynG= −247034

6829964.43∙ (−18.96 )=0.68 kN /cm2



σ nD=

M g

I io

∙ ynD= −247034

6829964.43∙133.44=−4.83 kN /cm2

13.1.2. Faza II – Uticaj pokretnog opterećenja

M p=−2301.72 kNm

σ bG=νb

M p

I io

∙ ybG=0.172 ∙

−2301726829964.43

∙ (−44.96 )=0.26kN /cm2

σ bD=νb

M p

Iio

∙ ybD=0.172∙

−2301726829964.43

∙ (−18.96 )=0.11 kN /cm2

σ nG=

M p

I io

∙ ynG= −230172

6829964.43∙ (−18.96 )=0.64kN /cm2

σ nD=

M p

I io

∙ ynD= −230172

6829964.43∙ 133.44=−4.50 kN /cm2



13.2. Proračun u t=t n

13.2.1. Uticaj sopstvene težine i tečenja

M φ=−2942.95 kNm

Nφ=809.36 kN

σ bG=νb φ( Nφ

A iφ

+M φ

Iiφ

∙ηG)−ρ ∙σ b 0=0.077 ∙( 809.361032.41

+ −2942955541235.85

∙ (−66.33 ))+0.56 ∙ 0.44=0.58kN /cm2

σ bD=νbφ ( N φ

A iφ

+M φ

I iφ

∙ ηD)−ρ∙ σ b 0=0.077 ( 809.361032.41

+ −2942955541235.85

∙ (−40.33 ))+0.56 ∙ 0.18=0.33 kN /cm2

σ nG=

N φ

A iφ

+M φ

I iφ

∙ ηG= 809.361032.41

+ −2942955541235.85

∙ (−40.33 )=2.93 kN /cm2

σ nD=

Nφ

Aiφ

+M φ

I iφ

∙ ηD= 890.361032.41

+ −2942955541235.85

∙112.07=−5.09 kN /cm2

13.2.2. Uticaj pokretnog opterećenja

M p=−2301.72 kNm

σ bG=νb

M p

I io

∙ ybG=0.172 ∙

−2301726829964.43

∙ (−44.96 )=0.26kN /cm2

σ bD=νb

M p

Iio

∙ ybD=0.172∙

−2301726829964.43

∙ (−18.96 )=0.11 kN /cm2

σ nG=

M p

I io

∙ ynG= −230172

6829964.43∙ (−18.96 )=0.64kN /cm2

σ nD=

M p

I io

∙ ynD= −230172

6829964.43∙ 133.44=−4.50 kN /cm2

13.2.3. Uticaj skupljanja betona

N sk=−3103.76 kN

M sk=−892.79 kNm

σ bG=νbφ( N sk

A iφ

+M sk

I iφ

∙ ηG)−εsk ∙ Ebφ=0.077 ∙(−3103.761032.41

+ −892795541235.85

∙ (−66.33 ))+0.0003 ∙1609=0.33 kN /cm2

σ bD=νbφ (N sk

A iφ

+M sk

I iφ

∙ ηD)−εsk ∙ Ebφ=0.077 ∙(−3103.761032.41

+ −892795541235.85

∙ (−40.33 ))+0.0003 ∙ 1609=0.30 kN /cm2



σ nG=

N sk

A iφ

+M sk

I iφ

∙ ηnG=−3103.76

1032.41+ −89279

5541235.85∙ (−40.33 )=−2.36 kN /cm2

σ nD=

N sk

Aiφ

+M sk

I iφ

∙ ηnD=−3103.76

1032.41+ −89279

5541235.85∙112.07=−4.81kN /cm2


RA RBRc


14. Moždanici

14.1. Proračun za statičsli sistem prosta greda sa privremenim osloncem u l /2

l=14,5 m l=14,5 m

14.1.1. Proračun smičućih sila

Presek nad osloncem

QA=RAIIa+RA

IIb+RAIIc=212.97+54.38+454.07=721.42 kN

Presek u l /2

Ql /2=RCIIa+RC

IIb+RCIIc=425.94+0+160.61=586.55 kN

T i=Q ∙ Sb

I io

Sb 0=νb ∙ Ab∙ yb 0=Abr ∙ yb0=1105.96 ∙ 23.91=26 443.50 cm3

Sb 2=νb ∙ Ab∙ yb 2=Abr ∙ yb2=1105.96 ∙ 36.13=39 958.33 cm3

T 0=QA ∙ Sb0

I io0 =721.42 ∙26443.50

3827999.46=4.98 kN /cm

T l/2=Ql/2 ∙ Sb 2

I io2 =586.55 ∙ 39958.33

6829964.43=3.43 kN /cm



T1 T2 T3

4.98

3.43

H1 H2 H3 H4

3.625 3.625 3.625 3.625

7.25

14.50

14.50 7.25

14.50

T 1=4.59 kN /cm

T 2=4.20 kN /cm

T 3=3.82 kN /cm

H 1=T 0+T 1

2∙ λ=4.98+4.59

2∙362.5=1734.56 kN

H 2=T 1+T 2

2∙ λ= 4.59+4.20

2∙362.5=1593.19 kN

H 3=T 2+T 3

2∙ λ=4.20+3.82

2∙ 362.5=1453.62 kN

H 4=T 3+T 4

2∙ λ=3.82+3.43

2∙362.5=1314.06 kN

14.1.2. Proračun nosivosti moždanika

I varijanta – Kombinovani moždanici (kruti + sidro)

Nd=RDE+Sa

RDE=A1 ∙ η∙ βRB

Sa=μ ∙ Aa ∙ σa

A1=10 ∙ 3=30 cm2

A2=b0 ∙ ( d p+hv )=28∙ (16+10 )=728 cm2

η=√ A2

A1

=√ 72830

=4.93>3 → η=3


h

1,2

310

13

26

m

10

1010

10

30


βRB=0.6 ∙ βb ( 28)=0.6 ∙ 5.0=3.0 kN /cm2

RDE=30 ∙3 ∙3.0=270 kN

μ=0.7−za zatvorena sidra

σ a=140 MPa=14 kN /cm2

Aa=r2 ∙ π

4=1.02 ∙ π

4=0.785 cm2

Sa=0.7 ∙ 0.785∙14=7.69 kN

Sa '=0.707 ∙ Sa=0.707 ∙7.69=5.43 kN

Nd=270+7.69=277.69 kN

II varijanta – Elastični moždanici (čep sa glavom)

30

26

8108

1,2

1:3

d=20 mm

1511

Usvojeno : d=20 mm, h=150 mm

d=20 mm≤ 2.5 ∙ t=2.5∙ 14=35 mm

hd=150

20=7.5>4

100 mm>3.5 ∙ d=3.5 ∙20=70 mm

Nosivost

Rd 1=0.25 ∙ d2 ∙√Eb ∙ βb (28 )



Rd 2=0.7 ∙d2 π

4σ n

Eb=36.21GPa=3621 kN /cm2

βb (28 )=5.0 kN /cm2

Rd 1=0.25 ∙2.02 ∙√3621 ∙ 5.0=134.55 kN

σ n=45 kN /cm2

Rd 2=0.7 ∙2.02 π

4∙ 45=98.96 kN

Ndop=2 ∙ min ( Rd 1 ;Rd 2 )=2∙min (134.55 ;98.96 )=2∙ 98.96=197.92 kN

14.1.3. Određivanje broja i razmaka moždanika duž raspona nosača

I varijanta

e=max { 60 cm3 ∙ d p=3 ∙16=48 cm

n1=H 1

Ndop

=1734.56277.69

=6.25 kom→ usvajam :n1=7 kom

e1=λn1

=362.57

=51.79 cm→ usvajam :e1=50 cm

n2=H 2

Ndop

=1593.19277.69

=5.74 kom →usvajam :n1=6 kom

e2=λn2

=362.56


n3=H 3

N dop

=1453.62277.69




e3=λn3

=362.56


n4=H 4

Ndop

=1314.06277.69

=4.73kom→ usvajam :n1=5kom

e4=λ

n4

=362.55


Usvajam razmak moždanika od e=50 cm što uz raspon od l=29 m daje ukupan broj krutih moždanika 58 kom.

II varijanta

n1=H 1

Ndop

=1734.56197.92

=8.76 kom →usvajam : n1=9 kom

e1=λn1

=362.59


n2=H 2

Ndop

=1593.19197.92


e2=λn2

=362.59

=40.28 cm →usvajam : e2=40 cm

n3=H 3

N dop

=1453.62197.92


e3=λn3

=362.58

=45.31 cm →usvajam : e3=45 cm

n4=H 4

Ndop

=1314.06197.92

=6.64 kom→ usvajam :n1=7kom

e3=λn3

=362.57


Usvajam razmak moždanika od e=40 cm što uz raspon od l=29 m daje ukupan broj krutih moždanika 73 kom.

Rekapitulacija

I varijanta

Prva i zadnja četvrtina

n=72550

=14.50 → usvojeno :n=15 kom

Druga i treća četvrtina



n=72560

=12.08 → usvojeno :n=13 kom

Ukupno

n=56 kom

II varijanta

Prva i zadnja četvrtina

n=72540

=18.12 → usvojeno: n=19 kom

Druga i treća četvrtina

n=72545

=16.11→ usvojeno :n=17 kom

Ukupno

n=72 kom


1. graficki rad

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