12 - Transverzalne Sile

8/20/2019 12 - Transverzalne Sile

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PRORAUN PRESEKA ZA GRANINE

UTICAJE TRANSVERZALNIH SILA• PRORAUN PREMA TEORIJI GRANINIH STANJA

•SAVIJANJE

AB PRESEKA POPRE

NIM SILAMAKONTROLA GLAVNIH NAPONA ZATEZANJA

2

2 b b

1 22 4

,

σ σσ = ± + τ


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d

y b 1

a 1

b

h

y b 2

h

- x

x

z b 1

G b

Aa1

D bu1

Zau

ε bd

ε b σ b

Mu

a 1

εa1

η 1 x

2

2 b b

1 22 4

,

σ σσ = ± + τ 1 2,σ = ±τ

2

T

b z,max max

min

−σ = τ =⋅

i

2

i

T S

b I,max max

min

⋅−σ = τ =

⋅


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( ) ( )11 b

b

uu

auumu ayz

dz

dNtg Ntgtg

h

MTT +−×−β×+β+α×= m

a1 b1 1y y a= −

( ) 1auu11 buuau y NMay NMM ×+=−×+=

( )umu u

MT T tg tg

h= × α + βm

( ) ( )1a b

u1 bu

uumu

yzdzdNtg

hy Ntgtg

hMTT −×−

β+×+β+α×= m


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BAB 87


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1.61.51.31.10.80.6τr [MPa]

605040302015MB

• Napon smicanja:

• Ra#unska #vrsto%a betona pri smicanju, τr :

• Mogu%i slu#ajevi:1) konstruktivna popre#na armatura

2) prora#unska popre#na armaturaza prihvatanje uticaja od dejstvatransverzalnih sila Tmu

mu

n b

b

Ty

b y z

( )

( )

τ =⋅

n r τ ≤ τ

n r τ > τ


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1) potrebna površina armature seodreuje na osnovu TRu

2) beton ne u!estvuje u prijemu

uticaja od transverzalnih sila

r n r 3τ < τ < τ

Ru mu buT T T= −

( ) bu r n1

T 3 b z

2

= ⋅ τ − τ ⋅ ⋅

r n r 3 5τ < τ < τ

Ru muT T=

buT 0=


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Ru

kuTZ

sin=

αku Ru

kuZ TZs z ctg ctg( )sin

′ = =θ + α α

s z ctg ctg( )sin= θ + α α


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• Redukuvana sila smicanja na jedinicu dužinenosa#a:

• Potrebna površina preseka popre#ne armaturena jedinicu dužine nosa#a:

ku Ru

ak

v v

Z TActg ctg( )sin

′′ = =σ σ θ + α α

Ru

Ru n

TT b

z= = τ ⋅


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• Ukupna rekukovana merodavna sila smicanja nadužini osiguranja (horizontalna sila veze):

• Ukupna potrebna površina preseka popre#nearmature:

x b x b

Ru

vu Ru

x a x a

TH T dx dx

z

= =

= =

= =∫ ∫

x b

Ru vu

ak

v vx a

T H1

A dxctg ctg z ctg( ) sin (cos sin )

=

== =σ θ + α α σ θ + α ⋅ α∫


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( ) uv

Ru)1(

u ectgsincos

1

m

b

a ×θ×α+α×σ×

τ×

=

( ) vRu

)1(

u

u ctgsincos b

am

e σ×θ×α+α×τ××

=


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• Pri uglu nagiba θ


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• Za θ


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g=40kN/m

P=240kN

Primer 1: (v idet i mater ijal “07” na http: imk sus ... ! !! )

Dimenzionisati nosa! sistema proste grede,!iji su optere$enje i popre!ni presek prikazani na skici.

MB 30, RA 400/500.


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g=40kN/m

A B

Ag=120 Bg=120

1 6 0

1 2 0

4 0

1 2 0 m

a x . M g = 1 8 0 k N m

Mg

Tg


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8 0

3

3 6

3 5 2

4 8 0

8 3 2

Mu

Tu m a x . M u = 8 3 2


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2RØ12

2RØ25

URØ10/25

3RØ25

3RØ25

2RØ25


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λ=308.4 91.6

88.5 88

τ ( 1 )

u , u

= 2 . 5

1

[MPa]

τAn =2.67

τARu=2.35

τCn=0.63τr =1.1

400

τ

τ ( 2 )

u , u

= 1 . 6

8

τ ( 3 )

u , u

= 1 . 0

1

131.9


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[MPa]

τBn=τBRu=3.81 τ

λ=200

τCn=2.79

τCRu=2.54Aτ

100.3 99.7

3τr

τ ( 1 )

u , u

= 1 . 6

8


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g = 30 kN/m MB 30

p = 20 kN/m RA 400/500


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336

420

252

672

378


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URØ8/20 4RØ22

5RØ22

2RØ12

B-B2RØ22

2RØ12

2RØ22

2RØ22

URØ10/20

4RØ22C-C


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τAn =2.01

τC,ln =2.52

τn


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kN5.657.47352

252.011.03T bu =××−×=

kN T T T bumu Ru

5.3545.65420 =−=−=

2RuRu cm/kN212.0

7.4735

5.354

z b

T=

×=

×=τ

( ) cm45.8a76.1011040212.035785.02e )1(uu =×=×+××××=

2l,C

n

l,C

mu cm/kN252.07.4735

420kN420T =

×=τ⇒=


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τn

τu,u


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( )θ×α+α×σ×××

=τ ctgsincose b

amv

u

)1(

u)1(

u,u

cm4.22102.035

785.02

b

ame

2

.min,uz

)1(

uu =××

×=

µ××

≤ −

cm20e.usv

cm25

cm5.262h

cm35 b

.mine uu =⇒

==

≤

( ) 2)1( u,u cm/kN090.0110402035785.02 =×+××××=τ

)1(


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cm11912.2

90.04.281

l,C

Ru

)1(

u,u

2 =×=ττ

×λ

MPa22.190.012.2)1( u,ul,C

max,Ru

l,C

Ru =−=τ−τ=τ∆

( )

2)1(

u cm07.120

110

1

402

122.035a =×

×+

×

×

×=

( ) uv

l,C

Ru)1(u e

ctgsincos1

m ba ×θ×α+α×σ× τ∆×=

cm4.1621194.2812 =−=λ∆

( )( ) 2)3( u,u cm/kN219.011040

2035

131.1785.02=×+××

×+×

=τ


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kN5.348354.1622

090.0212.0H k ,vu =×

×−=

( )θ×α+α×σ=

ctgsincos

HA

k k v

k ,vu

k ,a

( )2

k ,a cm16.60.1707.0707.040

5.348A =

×+×=

bAH k ,vu ×= τ


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τRu-τu,u


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τRu

-τu,u τRu

-τu,u

Ž Ž


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τRu

-τu,u

KOSIM PROFILIMA

DUŽINA OSIGURANJA DUŽINA OSIGURANJA

KOSIM PROFILIMA

τRu

-τu,u


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kN5.1077.47352

201.011.03TA bu =××

−×=

2A

n

A

mu cm/kN201.07.4735

336kN336T =

×

=τ⇒=

kN5.2285.107336TTT bumuRu =−=−=2

Ru cm/kN137.07.4735

5.228=

×=τ

( ) cm1.13a7.1611040137.035

785.02e )1(uu =×=×+×××

×=


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τAn =2.01

τARu=1.37

τ ( 1 )

u , u

= 0 . 9 0

τ ( 2 )

u , u

= 1 . 4 7

τn

τu,u

τRu

900)1(τ


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cm11937.1

90.04.181

A

Ru

)(

u,u

1 =×=ττ

×λ

MPa047.090.037.1)1( u,uA

max,Ru

A

Ru =−=τ−τ=τ∆

( )

2)1(

u cm412.020

110

1

402

047.035a =×

×+

×

×

×=

( ) uv

A

Ru)1(u e

ctgsincos1

m ba ×θ×α+α×σ× τ∆×=

cm4.621194.1811 =−=λ∆

( )( ) 2)2( u,u cm/kN147.011040

2035

503.0785.02=×+××

×+×

=τ


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τnτC,dn =1.51

τC,dRu =0.61


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kN5.1497.47352

151.011.03T d,C bu =××

−×=

2A

n

d,C

mu cm/kN151.07.4735

252kN252T =

×=τ⇒=

kN5.1025.149252TTT bumuRu =−=−=2

Ru cm/kN061.07.4735

5.102=

×=τ

( ) cm2.29a2.3711040061.035

785.02e )1(uu =×=×+×××

×=

12 - Transverzalne Sile

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