dimenzioniranje presjeka pbab-87 zad 1-19

Osnove betonskih konstrukcija Ak. 2012/2013

PBAB'87

PBAB '87

PRAVILNIK ZA BETON I ARMIRANI BETON

1987. godina


PBAB'87


PBAB 87

D I M E N Z I O N I R A NJ E P R E S J E K A

P B A B ' 87


PBAB 87

Osnove betonskih konstrukcija

PBAB '87 1

Zadatak 1.

Dimenzionirati AB presjek na osnovu zadanih podataka.

Statički utjecaji:

g p

g

p

M M 0

N 300 kNvlak

N 200 kN

= =

= ⎫⎪ −⎬= ⎪⎭

Presjek: kvadratni 40/40 cm

Materijal:

MB 30 → b 2 2

N kNf 20,5 2,05mm cm

= = - računska tlačna čvrstoća

RA 400/500 → av 2 2

N kNf 400 40mm cm

= = - granica razvlačenja

_______________________________________________________________

Traži se: pot Aa _______________________________________________________________

Pretpostavka: a 3ε ≥ ‰ → g p1,6 ; 1,8γ = γ =

Ultimna vlačna sila:

u g g p pN N N 1,6 300 1,8 200 840 kN= γ ⋅ + γ ⋅ = ⋅ + ⋅ =

Potrebna površina vlačne armature:

2ua,pot

av

N 840A 21,0 cmf 40

= = =

Usvojena armatura:

8 19φ ( )2aA 22,68 cm=

Raspored armature u poprečnom presjeku

1 8O19

2 O8

MB 30

RA 400/500

a = 2,5 cm


PBAB '87 2

Zadatak 2.


Statički utjecaji:

g

p

g

p

M 30 kNm

M 20 kNm

N 300 kNvlak

N 200 kN

=

=

= ⎫⎪ −⎬= ⎪⎭


Materijal:

MB 30 → b 2 2


= = - računska tlačna čvrstoća

RA 400/500 → av 2 2

N kNf 400 40mm cm

= = - granica razvlačenja

_______________________________________________________________ Ekscentricitet sile

M 30 20 d 40e 0,1m 10 cm 20 cmN 300 200 2 2

+= = = = < = =

+ (ekscentrični vlak, mali

ekscentricitet) Traži se: pot Aa1 i potAa2 _______________________________________________________________


Ultimne presječne sile:

u g g p p

u g g p p

M M M 1,6 30 1,8 20 84 kNm

N N N 1,6 300 1,8 200 840 kN

= γ ⋅ + γ ⋅ = ⋅ + ⋅ =

= γ ⋅ + γ ⋅ = ⋅ + ⋅ =

uu

u

M 84e 0,1m 10 cmN 840

= = = =

Zau1

Zau2

Tb

12

a1a2

Nu

Mu Zau1

Zau2

Nu

Tb Tb

Aa1

Aa22

1

u

Pretpostavka: 1 2d d 5 cm= =

a1 1d 40y d 5 15 cm2 2

= − = − =

a2 2d 40y d 5 15 cm2 2

= − = − =


PBAB '87 3

( ) ( )2

au1 a1 a2 u a2 u

a2 uau1 u

a1 a2

M 0

Z y y N y e

y e 15 10Z N 840 140 kNy y 15 15

=

⋅ + = ⋅ −

− −= ⋅ = ⋅ =

+ +

∑

( ) ( )1

au2 a1 a2 u a1 u

a1 uau2 u

a1 a2

M 0

Z y y N y e

y e 15 10Z N 840 700 kNy y 15 15

=

⋅ + = ⋅ +

+ += ⋅ = ⋅ =

+ +

∑

Kontrola:

au1 au2 u

H 0

Z Z N

140 700 840

840 840

=

+ =

+ =

=

∑


2au1a1,pot

av

Z 140A 3,50 cmf 40

= = =

2au2a2,pot

av

Z 700A 17,50 cmf 40

= = =

Usvojena armatura:

Aa1: 3 14φ ( )2aA 4,62 cm=

Aa2: 5 22φ ( )2aA 19,01cm=


1 5O22

3 O8/20

2 3O14

1 vil 1,pret

2 vil 2,pret

1,4d a 2,5 0,8 4 cm d 5 cm2 2

2,2d a 2,5 0,8 4,4 cm d 5 cm2 2

φ= + φ + = + + = ≠ =

φ= + φ + = + + = ≠ =

a1 1

a2 2

d 40y d 4 16 cm2 2d 40y d 4,4 15,6 cm2 2

= − = − =

= − = − =

MB 30

RA 400/500

a = 2,5 cm


PBAB '87 4

( ) ( )2

au1 a1 a2 u a2 u

a2 uau1 u

a1 a2

M 0

Z y y N y e

y e 15,6 10Z N 840 148,9 kNy y 16 15,6

=

⋅ + = ⋅ −

− −= ⋅ = ⋅ =

+ +

∑

( ) ( )1

au2 a1 a2 u a1 u

a1 uau2 u

a1 a2

M 0

Z y y N y e

y e 16 10Z N 840 691,1kNy y 16 15,6

=

⋅ + = ⋅ +

+ += ⋅ = ⋅ =

+ +

∑

Kontrola:

au1 au2 u

H 0

Z Z N

148,9 691,1 840

840 840

=

+ =

+ =

=

∑

2au1a1,pot

av

Z 148,9A 3,72 cmf 40

= = =

2au2a2,pot

av

Z 691,1A 17,28 cmf 40

= = =

Usvojena armatura:

Aa1: 3 14φ ( )2aA 4,62 cm=

Aa2: 5 22φ ( )2aA 19,01cm=


PBAB 87 5

Zadatak 3.


Statički utjecaji:

g

p

g p

M 80 kNm

M 50 kNm

N N 0

=

=

= =

Presjek: pravokutni 30/50 cm

Materijal:

a) MB 30 b) MB 30 c) MB 20

RA 400/500 GA 240/360 RA 400/500 _______________________________________________________________

Traži se:

a,potA i

b a au( 10ε ε = ε = ‰) ili a b bu( 3,5ε ε = ε = ‰) _______________________________________________________________

a) MB 30 RA 400/500


Ultimni moment:

u g g p pM M M 1,6 80 1,8 50 218 kNm= γ ⋅ + γ ⋅ = ⋅ + ⋅ =

Dimenzioniranje pomoću ''mau'' tabela

Ultimni moment u težištu vlačne armature:

( )au u uM M 218 kNm N 0= = =

Pretpostavka: d2,pret = 5 cm – udaljenost težišta vlačne armature do vlačnog ruba

pret 2,preth d d 50 5 45cm= − = − = - pretpostavljena statička visina presjeka

2au

au au2 2b

M 218 10m 0,175 m 0,338b h f 30 45 2,05

∗⋅= = = < =

⋅ ⋅ ⋅ ⋅ - 1-struko armiran presjek

mau = 0,175 → ωMa = 0,195

εa = εau = 10 ‰

εb = 3,24 ‰ (područje 2b)


2ba,pot Ma

av

f 2,05A b h 0,195 30 45 13,49 cmf 40

= ω ⋅ ⋅ ⋅ = ⋅ ⋅ ⋅ =

Usvojena armatura:

5 19φ ( )2aA 14,18 cm=


PBAB 87 6


1 5O19

2 2O12

3 O8/20

2 vil

2 pret

1,9d a 2,5 0,8 4,25 cm2 2

h d d 50 4,25 45,75 cm h 45 cm

φ= + φ + = + + =

= − = − = > =

b) MB 30 GA 240/360


u g g p pM M M 1,6 80 1,8 50 218 kNm= γ ⋅ + γ ⋅ = ⋅ + ⋅ =


( )au u uM M 218 kNm N 0= = =



2au

au au2 2b

M 218 10m 0,175 m 0,338b h f 30 45 2,05

∗⋅= = = < =


mau = 0,175 → ωMa = 0,195

εa = εau = 10 ‰



2ba,pot Ma

av

f 2,05A b h 0,195 30 45 22,49 cmf 24

= ω ⋅ ⋅ ⋅ = ⋅ ⋅ ⋅ =

Usvojena armatura:

6 22φ ( )2aA 22,81cm=

MB 30

RA 400/500

a = 2,5 cm


PBAB 87 7


1 6O22

2 2O12

3 O8/20

2 vil

2 pret

d a 1,0 2,5 0,8 2,2 1,0 6,5 cm

h d d 50 6,5 43,5 cm h

= + φ + φ + = + + + =

= − = − = <

Ponovljeni proračun Pretpostavka: d2,pret = 8 cm

pret 2h d d 50 8 42cm= − = − =

2au

au au2 2b

M 218 10m 0,200 m 0,338b h f 30 42 2,05

∗⋅= = = < =


mau = 0,200 → ωMa = 0,226

εa = 9,01 ‰

εb = εbu = 3,50 ‰ (područje 3a)


2ba,pot Ma

av

f 2,05A b h 0,226 30 42 24,32 cmf 24

= ω ⋅ ⋅ ⋅ = ⋅ ⋅ ⋅ =

Usvojena armatura:

5 25φ ( )2aA 24,54 cm=


1 5O25

2 2O12

3 O8/20

MB 30

GA 240/360

a = 2,5 cm

MB 30

GA 240/360

a = 2,5 cm


PBAB 87 8

2 vil

2 pret

d a 1,5 2,5 0,8 2,5 1,5 7,3 cm

h d d 50 7,3 42,7 cm h

= + φ + φ + = + + + =

= − = − = >

c) MB 20 RA 400/500


u g g p pM M M 1,6 80 1,8 50 218 kNm= γ ⋅ + γ ⋅ = ⋅ + ⋅ =


( )au u uM M 218 kNm N 0= = =


pret 2,preth d d 50 5 45cm= − = − = - pretpostavlja statička visina presjeka

2au

au au2 2b

M 218 10m 0,256 m 0,338b h f 30 45 1,4

∗⋅= = = < =


mau = 0,256 → ωMa = 0,304

εa = 5,84 ‰



2ba,pot Ma

av

f 1,4A b h 0,304 30 45 14,36 cmf 40

= ω ⋅ ⋅ ⋅ = ⋅ ⋅ ⋅ =

Usvojena armatura:

4 22φ ( )2aA 15,21cm=


1 4O22

2 2O12

3 O8/20

2 vil

2 pret

2,2d a 2,5 0,8 4,4 cm2 2

h d d 50 4,4 45,6 cm h

φ= + φ + = + + =

= − = − = >

MB 20

RA 400/500

a = 2,5 cm


PBAB 87 9

Zadatak 4.


Statički utjecaji:

g

p

g

p

M 80 kNm

M 50 kNm

N 150 kNtlak

N 100 kN

=

=

= − ⎫⎪ −⎬= − ⎪⎭


Materijal:

MB 30

RA 400/500 _______________________________________________________

Traži se:

a,potA i

b a au( 10ε ε = ε = ‰) ili a b bu( 3,5ε ε = ε = ‰) _______________________________________________________



u g g p pM M M 1,6 80 1,8 50 218 kNm= γ ⋅ + γ ⋅ = ⋅ + ⋅ =

u g g p pN N N 1,6 ( 150) 1,8 ( 100) 420 kN= γ ⋅ + γ ⋅ = ⋅ − + ⋅ − = −


Pretpostavka: d2,pret = 5 cm

Udaljenost težišta vlačne armature od težišta presjeka:

a 2,pretd 50y d 5 20 cm2 2

= − = − =

Ultimni moment vanjskih sila u težištu vlačne armature:

au u u aM M N y 218 ( 420) 0,20 302 kNm= − ⋅ = − − ⋅ =

2au

au au2 2b

M 302 10m 0,243 m 0,338b h f 30 45 2,05

∗⋅= = = < =


mau = 0,243 → ωMa = 0,285

εa = 6,45 ‰



PBAB 87 10


2b ua,pot Ma

av av

f N 2,05 420A b h 0,285 30 45 19,72 10,5 9,22 cmf f 40 40

−= ω ⋅ ⋅ ⋅ + = ⋅ ⋅ ⋅ + = − =

Usvojena armatura:

5 16φ ( )2aA 10,05 cm=


1 5O16

2 2O12

3 O8/20

2 vil

pret

1,6d a 2,5 0,8 4,1cm2 2

h 50 4,1 45,9 cm h

φ= + φ + = + + =

= − = >

MB 30

RA 400/500

a = 2,5 cm


PBAB 87 11

Zadatak 5.


Statički utjecaji:

g

p

g

p

M 80 kNm

M 50 kNm

N 150 kNvlak

N 100 kN

=

=

= ⎫⎪ −⎬= ⎪⎭


Materijal:

MB 30

RA 400/500 _______________________________________________________________

Traži se:

a,potA i




u g g p pM M M 1,6 80 1,8 50 218 kNm= γ ⋅ + γ ⋅ = ⋅ + ⋅ =

u g g p pN N N 1,6 150 1,8 100 420 kN= γ ⋅ + γ ⋅ = ⋅ + ⋅ =




a 2,pretd 50y d 8 17 cm2 2

= − = − =


au u u aM M N y 218 420 0,17 146,6 kNm= − ⋅ = − ⋅ =


2au

au au2 2b

M 146,6 10m 0,135 m 0,338b h f 30 42 2,05

∗⋅= = = < =


mau = 0,135 → ωMa = 0,146

εa = εau = 10 ‰



PBAB 87 12


2b ua,pot Ma

av av

f N 2,05 420A b h 0,146 30 42 9,43 10,5 19,93 cmf f 40 40

= ω ⋅ ⋅ ⋅ + = ⋅ ⋅ ⋅ + = + =

Usvojena armatura:

3 25 2 19φ + φ ( )2aA 14,73 5,67 20,4 cm= + =


1 3O25

3 2O12

4 O8/20

2 2O19

2 vil

2 pret

d a 1,0 2,5 0,8 2,5 1,0 6,8 cm

h d d 50 6,8 43,2 h

= + φ + φ + = + + + =

= − = − = >

MB 30

RA 400/500

a = 2,5 cm


PBAB 87 13

Zadatak 6.


Statički utjecaji:

g

p

g p

M 80 kNm

M 50 kNm

N N 0

=

=

= =


Materijal:

MB 30

RA 400/500 _______________________________________________________________

Traži se:

a,potA i



Ultimni moment:

u g g p pM M M 1,6 80 1,8 50 218 kNm= γ ⋅ + γ ⋅ = ⋅ + ⋅ =



( )au u uM M 218 kNm N 0= = =


pret 2,preth d d 40 7 33cm= − = − =

2au

au au2 2b

M 218,0 10m 0,390 m 0,338b h f 25 33 2,05

∗⋅= = = > =


_______________________________________________________________

Traži se:

a,potA i 'a,potA

b bu( 3,5ε = ε = ‰ ; a 3,0ε = ‰) _______________________________________________________________

Dimenzioniranje 2-struko armiranih presjeka pomoću ''mau'' tabela

Najveći moment koji može preuzeti 1-struko armiran presjek u stanju granične

ravnoteže: 2 2

au au bM m b h f 0,338 0,25 33 2,05 188,6 kNm∗ ∗= ⋅ ⋅ ⋅ = ⋅ ⋅ ⋅ =


PBAB 87 14

au au auM M M 218 188,6 29,4 kNm∗Δ = − = − = - razlika momenata koju uravnotežuje

spreg sila u tlačnoj i dodatnoj vlačnoj armaturi.


( )

( )

au aua,pot a,pot a

1 avz av

2

M MA A A

h d fk h f

188,6 29,4 18,41 2,53 21,0 cm0,776 0,33 40 0,33 0,04 40

∗∗

∗

Δ= + Δ = + =

− ⋅⋅ ⋅

= + = + =⋅ ⋅ − ⋅

Pretpostavljeno: d1,pret = 4 cm

Potrebna površina tlačne armature:

( ) ( )' 2aua,pot

1 av

M 29,4A 2,53 cmh d f 0,33 0,04 40Δ

= = =− ⋅ − ⋅

'a,pot a,potA A>

Usvojena armatura:

Vlačna - 3 25 2 22φ + φ ( )2aA 14,73 7,60 22,33 cm= + =

Tlačna - 3 12φ ( )2aA 3,39 cm=

Dimenzioniranje 2-struko armiranih presjeka bez uporabe tabela

Traži se:

a,potA i 'a,potA

b bu( 3,5ε = ε = ‰ ; a 3,0ε = ‰)


Ultimni moment:

u g g p pM M M 1,6 80 1,8 50 218 kNm= γ ⋅ + γ ⋅ = ⋅ + ⋅ =


( )au u uM M 218 kNm N 0= = =


ravnoteže: ∗ ∗ ∗

∗ ∗

∗ ∗

∗ ∗ ∗ ∗ ∗

= ⋅

= α ⋅ ⋅ ⋅ ⋅

= ⋅

= α ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ = ⋅ ⋅ ⋅ α ⋅ ⋅

au bu

bu x b

z2

au x b z b x z

M P z

P k b h f

z k h

M k b h f k h b h f k k


PBAB 87 15

Koeficijent punoće presjeka:

b 2ε ≥ ‰ → b

b

3 2 3 3,5 2 0,80953 3 3,5⋅ ε − ⋅ −

α = = =⋅ ε ⋅

b = 0,25 m


pret 2,preth d d 40 7 33cm= − = − =

Tlačna računska čvrstoća betona:

b 2 2


= =

Koeficijent položaja neutralne osi:

bx

b a

3,5k 0,5383,5 3,0

∗ ε= = =ε + ε +

Koeficijent kraka unutrašnjih sila:

z p xk 1 k k 1 0,416 0,538 0,776∗ ∗ ∗= − ⋅ = − ⋅ =

b 2ε ≥ ‰ → ( ) ( )

2 2b b

pb b

3 4 2 3 3,5 4 3,5 2k 0,4162 3 2 2 3,5 3 3,5 2

∗ ⋅ ε − ⋅ ε + ⋅ − ⋅ += = =

⋅ ε ⋅ ⋅ ε − ⋅ ⋅ ⋅ −

2auM 0,25 33 2,05 0,8095 0,538 0,776 188,6 kNm∗ = ⋅ ⋅ ⋅ ⋅ ⋅ =

au auM 188,6 kNm M 218 kNm∗ = < = → 2-struko armiran presjek

au au auM M M 218 188,6 29,4 kNm∗Δ = − = − = - razlika momenata koju uravnotežuje



( )

( )

au aua,pot a,pot a

1 avz av

2

M MA A Ah d fk h f

188,6 29,4 18,41 2,53 21,0 cm0,776 0,33 40 0,33 0,04 40

∗∗

∗

Δ= + Δ = + =

− ⋅⋅ ⋅

= + = + =⋅ ⋅ − ⋅

Pretpostavljeno: d1 = 4 cm


( ) ( )' 2aua,pot

1 av

M 29,4A 2,53 cmh d f 0,33 0,04 40Δ

= = =− ⋅ − ⋅

'a,pot a,potA A>

Usvojena armatura:

Vlačna - 3 25 2 22φ + φ ( )2aA 14,73 7,60 22,33 cm= + =

Tlačna - 3 12φ ( )2aA 3,39 cm=


PBAB 87 16


1 3O25

3 3O12

4 O8/20

2 2O22

2 vil

2 pret

1 vil

d a 1,0 2,5 0,8 2,5 1,0 6,8 cm

h d d 40 6,8 33,2 h

12d a 2,5 0,8 3,9 4,0 cm2 2

= + φ + φ + = + + + =

= − = − = >

φ= + φ + = + + = ∼

MB 30

RA 400/500

a = 2,5 cm


PBAB 87 17

Zadatak 7.


Statički utjecaji:

g

p

g

p

M 100 kNm

M 80 kNm

N 80 kNtlak

N 60 kN

=

=

= − ⎫⎪ −⎬= − ⎪⎭


Materijal:

MB 40

RA 400/500 _______________________________________________________________

Traži se:

a,potA i




u g g p pM M M 1,6 100 1,8 80 304 kNm= γ ⋅ + γ ⋅ = ⋅ + ⋅ =

u g g p pN N N 1,6 ( 80) 1,8 ( 60) 236 kNm= γ ⋅ + γ ⋅ = ⋅ − + ⋅ − = −



a 2,pretd 40y d 7 13 cm2 2

= − = − =


au u u aM M N y 304 ( 236) 0,13 304 236 0,13 334,7 kNm= − ⋅ = − − ⋅ = + ⋅ =

pret 2,preth d d 40 7 33cm= − = − =

2au

au au2 2b

M 334,7 10m 0,402 m 0,338b h f 30 33 2,55

∗⋅= = = > =


Traži se:

a,potA i 'a,potA

b bu 3,5ε = ε = ‰ ; a 3ε = ‰


PBAB 87 18



ravnoteže: 2 2

au au bM m b h f 0,338 0,30 33 2,55 281,6 kNm∗ ∗= ⋅ ⋅ ⋅ = ⋅ ⋅ ⋅ =

au au auM M M 334,7 281,6 53,1kNm∗Δ = − = − = - razlika momenata koju uravnotežuje



( )

( )

au u aua,pot a,pot a

av 1 avz av

2

M N MA A Af h d fk h f

281,6 236 53,1 27,49 5,9 4,74 26,33 cm0,776 0,33 40 40 0,33 0,05 40

∗∗

∗

Δ= + Δ = + +

− ⋅⋅ ⋅

−= + + = − + =

⋅ ⋅ − ⋅

Pretpostavljeno: d1 = 5 cm


( ) ( )' 2aua,pot

1 av

M 53,1A 4,74 cmh d f 0,33 0,05 40Δ

= = =− ⋅ − ⋅

'a,pot a,potA A>

Usvojena armatura:

Vlačna - 4 25 2 22φ + φ ( )2aA 19,63 7,60 27,23 cm= + =

Tlačna - 4 14φ ( )' 2aA 6,16 cm=


Traži se:

a,potA i 'a,potA

b bu 3,5ε = ε = ‰ ; a 3ε = ‰



u g g p pM M M 1,6 100 1,8 80 304 kNm= γ ⋅ + γ ⋅ = ⋅ + ⋅ =

u g g p pN N N 1,6 ( 80) 1,8 ( 60) 236 kNm= γ ⋅ + γ ⋅ = ⋅ − + ⋅ − = −


d 40y d 7 13 cma 22 2= − = − =


au u u aM M N y 304 ( 236) 0,13 304 236 0,13 334,7 kNm= − ⋅ = − − ⋅ = + ⋅ =


PBAB 87 19


ravnoteže

au bu

bu x b

z

2au x b z b x z

M P z

P k b h f

z k h

M k b h f k h b h f k k

∗ ∗ ∗

∗ ∗

∗ ∗

∗ ∗ ∗ ∗ ∗

= ⋅

= α ⋅ ⋅ ⋅ ⋅

= ⋅

= α ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ = ⋅ ⋅ ⋅ α ⋅ ⋅


b 2ε ≥ ‰ → b

b

3 2 3 3,5 2 0,80953 3 3,5⋅ ε − ⋅ −

α = = =⋅ ε ⋅

b = 0,30 m

pret 2,preth d d 40 7 33cm= − = − =


b 2

kNf 25,5 MPa 2,55cm

= =


bx

b a

3,5k 0,5383,5 3,0

∗ ε= = =ε + ε +


z p xk 1 k k 1 0,416 0,538 0,776∗ ∗ ∗= − ⋅ = − ⋅ =

b 2ε ≥ ‰ → ( ) ( )

2 2b b

pb b

3 4 2 3 3,5 4 3,5 2k 0,4162 3 2 2 3,5 3 3,5 2

∗ ⋅ ε − ⋅ ε + ⋅ − ⋅ += = =

⋅ ε ⋅ ⋅ ε − ⋅ ⋅ ⋅ −

2auM 0,30 33 2,55 0,8095 0,538 0,776 281,5 kNm∗ = ⋅ ⋅ ⋅ ⋅ ⋅ =

au auM 281,5 kNm M 334,7 kNm∗ = < = → 2-struko armiran presjek

au au auM M M 334,7 281,5 53,2 kNm∗Δ = − = − = - razlika momenata koju uravnotežuje



( )

( )


av 1 avz av

2


281,5 236 53,2 27,48 5,9 4,75 26,33 cm0,776 0,33 40 40 0,33 0,05 40

∗∗

∗

Δ= + Δ = + +

− ⋅⋅ ⋅

−= + + = − + =

⋅ ⋅ − ⋅



PBAB 87 20


( ) ( )' 2aua,pot

1 av

M 53,2A 4,75 cmh d f 0,33 0,05 40Δ

= = =− ⋅ − ⋅

'a,pot a,potA A>

Usvojena armatura:

Vlačna - 4 25 2 22φ + φ ( )2aA 19,63 7,60 27,23 cm= + =

Tlačna - 4 14φ ( )' 2aA 6,16 cm=


1 4O25

3 4O14

4 O8/20

2 2O22

2 vil

2 pret

1 vil 1,pret

d a 1,0 2,5 0,8 2,5 1,0 6,8 cm

h d d 40 6,8 33,2 h

14d a 2,5 0,8 4,0 d2 2

= + φ + φ + = + + + =

= − = − = >

φ= + φ + = + + = <

MB 30

RA 400/500

a = 2,5 cm


PBAB 87 21

Zadatak 8.


Statički utjecaji:

g

p

g

p

M 100 kNm

M 80 kNm

N 80 kNvlak

N 60 kN

=

=

= ⎫⎪ −⎬= ⎪⎭


Materijal:

MB 40

RA 400/500 _______________________________________________________________

Traži se:

a,potA i




u g g p pM M M 1,6 100 1,8 80 304 kNm= γ ⋅ + γ ⋅ = ⋅ + ⋅ =

u g g p pN N N 1,6 80 1,8 60 236 kNm= γ ⋅ + γ ⋅ = ⋅ + ⋅ =



a 2,pretd 40y d 8 12 cm2 2

= − = − =

pret 2,preth d d 40 8 32 cm= − = − =


au u u aM M N y 304 236 0,12 275,7 kNm= − ⋅ = − ⋅ =

2au

au au2 2b

M 275,7 10m 0,352 m 0,338b h f 30 32 2,55

∗⋅= = = > =


Traži se:

a,potA i 'a,potA

b bu 3,5ε = ε = ‰ ; a 3ε = ‰


PBAB 87 22



ravnoteže: 2 2

au au bM m b h f 0,338 0,30 32 2,55 264,8 kNm∗ ∗= ⋅ ⋅ ⋅ = ⋅ ⋅ ⋅ =


au au auM M M 275,7 264,8 10,9 kNm∗Δ = − = − = - razlika momenata koju uravnotežuje spreg sila u tlačnoj i dodatnoj vlačnoj armaturi.


( )

( )


av 1 avz av

2

M N MA A A

f h d fk h f

264,8 236 10,9 26,66 5,9 0,97 33,53 cm0,776 0,32 40 40 0,32 0,04 40

∗∗

∗

Δ= + Δ = + + =

− ⋅⋅ ⋅

= + + = + + =⋅ ⋅ − ⋅



( ) ( )' 2aua,pot

1 av

M 10,9A 0,97 cmh d f 0,32 0,04 40Δ

= = =− ⋅ − ⋅

'a,pot a,potA A>

Usvojena armatura:

Vlačna - 4 25 4 22φ + φ ( )2aA 19,63 15,20 34,83 cm= + =

Tlačna - 2 12φ ( )' 2aA 2,26 cm=


Traži se:

a,potA i 'a,potA

b bu 3,5ε = ε = ‰ ; a 3ε = ‰



u g g p pM M M 1,6 100 1,8 80 304 kNm= γ ⋅ + γ ⋅ = ⋅ + ⋅ =

u g g p pN N N 1,6 ( 80) 1,8 ( 60) 236 kNm= γ ⋅ + γ ⋅ = ⋅ − + ⋅ − = −


a 2d 40y d 8 12 cm2 2

= − = − =


PBAB 87 23

pret 2,preth d d 40 8 32 cm= − = − =


au u u aM M N y 304 236 0,12 275,7 kNm= − ⋅ = − ⋅ =


ravnoteže

au bu

bu x b

z

M P z

P k b h f

z k h

∗ ∗ ∗

∗ ∗

∗ ∗

= ⋅

= α ⋅ ⋅ ⋅ ⋅

= ⋅

2au x b z b x zM k b h f k h b h f k k∗ ∗ ∗ ∗ ∗= α ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ = ⋅ ⋅ ⋅ α ⋅ ⋅


b 2ε ≥ ‰ → b

b

3 2 3 3,5 2 0,80953 3 3,5⋅ ε − ⋅ −

α = = =⋅ ε ⋅

b = 0,30 m


b 2 2


= =


bx

b a

3,5k 0,5383,5 3,0

∗ ε= = =ε + ε +


z p xk 1 k k 1 0,416 0,538 0,776∗ ∗ ∗= − ⋅ = − ⋅ =

b 2ε ≥ ‰ → ( ) ( )

2 2b b

pb b

3 4 2 3 3,5 4 3,5 2k 0,4162 3 2 2 3,5 3 3,5 2

∗ ⋅ ε − ⋅ ε + ⋅ − ⋅ += = =

⋅ ε ⋅ ⋅ ε − ⋅ ⋅ ⋅ −

2auM 0,30 32 2,55 0,8095 0,538 0,776 264,7 kNm∗ = ⋅ ⋅ ⋅ ⋅ ⋅ =




( )

( )


av 1 avz av

2

M N MA A A

f h d fk h f

264,7 236 11,0 26,65 5,9 0,98 33,53 cm0,776 0,32 40 40 0,32 0,04 40

∗∗

∗

Δ= + Δ = + + =

− ⋅⋅ ⋅

= + + = + + =⋅ ⋅ − ⋅



PBAB 87 24


( ) ( )' 2aua,pot

1 av

M 11A 0,98 cmh d f 0,32 0,04 40Δ

= = =− ⋅ − ⋅

'a,pot a,potA A>

Usvojena armatura:

Vlačna - 4 25 4 22φ + φ ( )2aA 19,63 15,20 34,83 cm= + =

Tlačna - 2 12φ ( )' 2aA 2,26 cm=


1 4O25

3 2O12

4 O8/20

2 4O22

1 vil 11,2d a 2,5 0,8 3,9 cm pret d 4 cm

2 2φ

= + φ + = + + = < =

2 vil

2 pret

d a 1,0 2,5 0,8 2,5 1,5 7,3 cm d d

h d d 40 7,3 32,7 cm h

= + φ + φ + = + + + = −

= − = − = >

MB 30

RA 400/500

a = 2,5 cm


PBAB 87 25

Zadatak 9.


Statički utjecaji:

g

p

g

p

M 300 kNm

M 250 kNm

N 760 kNtlak

N 540 kN

=

=

= − ⎫⎪ −⎬= − ⎪⎭


Materijal:

MB 30

RA 400/500 _______________________________________________________________

Traži se:

a,potA i




u g g p pM M M 1,6 300 1,8 250 930 kNm= γ ⋅ + γ ⋅ = ⋅ + ⋅ =

( ) ( )u g g p pN N N 1,6 760 1,8 540 2188 kNm= γ ⋅ + γ ⋅ = ⋅ − + ⋅ − = −



a 2d 80y d 7 33 cm2 2

= − = − =


( )au u u aM M N y 930 2188 0,33 1652 kNm= − ⋅ = − − ⋅ =


Pretpostavljena statička visina presjeka:

pret 2,preth d d 80 7 73 cm= − = − =

2au

au au2 2b

M 1652 10m 0,504 m 0,338b h f 30 73 2,05

∗⋅= = = > =


Traži se:

a,potA i 'a,potA

( a 3ε = ‰ i b bu 3,5ε = ε = ‰)


PBAB 87 26


ravnoteže: 2 2

au au bM m b h f 0,338 0,30 73 2,05 1107,7 kNm∗ ∗= ⋅ ⋅ ⋅ = ⋅ ⋅ ⋅ =




( )

( )


av 1 avz av

2


1107,7 ( 2188) 544,3 48,89 54,7 20,62 14,81cm0,776 0,73 40 40 0,73 0,07 40

∗∗

∗

Δ= + Δ = + + =

− ⋅⋅ ⋅

−= + + = − + =

⋅ ⋅ − ⋅



( ) ( )' 2aua,pot

1 av

M 544,7A 20,62 cmh d f 0,73 0,07 40Δ

= = =− ⋅ − ⋅

'a,pot a,potA A< → simetrično armiranje presjeka

Traži se:

a1,potA = a2,potA ( )a1 a2A A= i

a b bu( 3,5ε ε = ε = ‰)

Dimenzioniranje pomoću interakcionih dijagrama (''mu - nu'' postupak)

Pretpostavka: a 0 ε < ‰ → g p1,9 ; 2,1γ = γ =

u g g p pM M M 1,9 300 2,1 250 1095 kNm= γ ⋅ + γ ⋅ = ⋅ + ⋅ =

( ) ( )u g g p pN N N 1,9 760 2,1 540 2578 kNm= γ ⋅ + γ ⋅ = ⋅ − + ⋅ − = −

1 2d d 7 0,0875d d 80= = =

uu 2 2

b

uu

b

M 1095Relativni moment m 0,278b d f 0,3 80 2,05

N 2578Relativna normalna sila n 0,524b d f 30 80 2,05

⎫= = = ⎪⋅ ⋅ ⋅ ⋅ ⎪⎬⎪= = = ⎪⋅ ⋅ ⋅ ⋅ ⎭

→ a 1,7ε ∼ ‰ >0 ‰


PBAB 87 27

1 2

u

u

d d 0,075d d

m 0,278

n 0,524

⎫= = ⎪⎪⎪= ⎬⎪

= ⎪⎪⎭

→ a 1,8ε ∼ ‰ → 0,395μ =

1 2

u

u

d d 0,100d d

m 0,278

n 0,524

⎫= = ⎪⎪⎪= ⎬⎪

= ⎪⎪⎭

→ a 1,8ε ∼ ‰ → 0,42μ =

a1,8 1,8 1,8 0

2+

ε = >∼ ‰

Pretpostavka: a 2,0 ε = ‰ → g p1,7 ; 1,9γ = γ =

u g g p pM M M 1,7 300 1,9 250 985 kNm= γ ⋅ + γ ⋅ = ⋅ + ⋅ =

( ) ( )u g g p pN N N 1,7 760 1,9 540 2318 kNm= γ ⋅ + γ ⋅ = ⋅ − + ⋅ − = −

1 2

u

u

d d 0,075d d

m 0,25

n 0,471

⎫= = ⎪⎪⎪= ⎬⎪

= ⎪⎪⎭

→ a 2,0ε ∼ ‰ → 0,305μ =

1 2

u

u

d d 0,100d d

m 0,25

n 0,471

⎫= = ⎪⎪⎪= ⎬⎪

= ⎪⎪⎭

→ a 2,1ε ∼ ‰ → 0,32μ =

a a,pret2,0 2,1 2,05

2+

ε = ε∼ ∼

0,305 0,32 0,31252+

μ = =

2ba,pot

av

f 20,5A b d 0,3125 30 80 38,44 cmf 400

= μ ⋅ ⋅ ⋅ = ⋅ ⋅ ⋅ =

2a1,pot a2,.pot a,pot

1 1A A A 38,44 19,22 cm2 2

= = ⋅ = ⋅ =

Usvojena armatura: 6 22± φ ( )2a1 a2A A 22,81cm= =


PBAB 87 28


1 6O22

2 2O12

3 O8/20

1 6O22

1 2 vil 1d d a 1,0 2,5 0,8 2,2 1,0 6,5 cm pret d 7 cm= = + φ + φ + = + + + = < =

MB 30

RA 400/500

a = 2,5 cm


PBAB 87 29

Zadatak 10.


Statički utjecaji:

g

p

g

p

M 300 kNm

M 250 kNm

N 1000 kNtlak

N 800 kN

=

=

= − ⎫⎪ −⎬= − ⎪⎭


Materijal:

MB 40

RA 400/500 _______________________________________________________________

Traži se:

a,potA i




u g g p pM M M 1,6 300 1,8 250 930 kNm= γ ⋅ + γ ⋅ = ⋅ + ⋅ =

( ) ( )u g g p pN N N 1,6 1000 1,8 800 3040 kNm= γ ⋅ + γ ⋅ = ⋅ − + ⋅ − = −

Pretpostavka: d2 = 7 cm


a 2d 80y d 7 33 cm2 2

= − = − =


( )au u u aM M N y 930 3040 0,33 1933,2 kNm= − ⋅ = − − ⋅ =


Pretpostavljena statička visina presjeka:

pret 2,preth d d 80 7 73 cm= − = − =

2au

au au2 2b

M 1933,2 10m 0,474 m 0,338b h f 30 73 2,55

∗⋅= = = > =


Traži se:

a,potA i 'a,potA

( a 3ε = ‰ i b bu 3,5ε = ε = ‰)


PBAB 87 30


ravnoteže: 2 2

au au bM m b h f 0,338 0,30 73 2,55 1377,9 kNm∗ ∗= ⋅ ⋅ ⋅ = ⋅ ⋅ ⋅ =




( )

( )


av 1 avz av

2


1377,9 ( 3040) 555,3 60,81 76,0 21,03 5,84 cm0,776 0,73 40 40 0,73 0,07 40

∗∗

∗

Δ= + Δ = + + =

− ⋅⋅ ⋅

−= + + = − + =

⋅ ⋅ − ⋅



( ) ( )' 2aua,pot

1 av

M 555,3A 21,03 cmh d f 0,73 0,07 40Δ

= = =− ⋅ − ⋅

'a,pot a,potA A< → simetrično armiranje presjeka

Traži se:

a1,potA = a2,potA ( )a1 a2A A= i

a b bu( 3,5ε ε = ε = ‰)

Dimenzioniranje pomoću interakcionih dijagrama (''mu - nu'' postupak)

Pretpostavka: a 0ε < ‰ → g p1,9 ; 2,1γ = γ =

u g g p pM M M 1,9 300 2,1 250 1095 kNm= γ ⋅ + γ ⋅ = ⋅ + ⋅ =

( ) ( )u g g p pN N N 1,9 1000 2,1 800 3580 kN= γ ⋅ + γ ⋅ = ⋅ − + ⋅ − = −

Pretpostavka 1 2d d 7 cm= =

1 2d d 7 0,0875d d 80= = =

uu 2 2

b

uu

b

M 1095Relativni moment m 0,224b d f 0,3 80 2,55

N 3580Relativna normalna sila n 0,585b d f 30 80 2,55

⎫= = = ⎪⋅ ⋅ ⋅ ⋅ ⎪⎬⎪= = = ⎪⋅ ⋅ ⋅ ⋅ ⎭


PBAB 87 31

1 2

u

u

d d 0,075d d

m 0,224n 0,585

⎫= = ⎪⎪

= ⎬⎪= ⎪⎭

→ a 1,4ε ∼ ‰ → 0,285μ =

1 2

u

u

d d 0,100d d

m 0,224n 0,585

⎫= = ⎪⎪

= ⎬⎪= ⎪⎭

→ a 1,3ε ∼ ‰ → 0,31μ =

a1,4 1,3 1,35 0

2+

ε = >∼ ‰

Pretpostavka: a 1,5 ε = ‰ → g p1,75 ; 1,95γ = γ =

u g g p pM M M 1,75 300 1,95 250 1012,5 kNm= γ ⋅ + γ ⋅ = ⋅ + ⋅ =

( ) ( )u g g p pN N N 1,75 1000 1,95 800 3310 kNm= γ ⋅ + γ ⋅ = ⋅ − + ⋅ − = −

uu 2 2

b

uu

b

M 1012,5m 0,207b d f 0,3 80 2,55

N 3310n 0,541b d f 30 80 2,55

⎫= = = ⎪⋅ ⋅ ⋅ ⋅ ⎪⎬⎪= = = ⎪⋅ ⋅ ⋅ ⋅ ⎭

1 2

u

u

d d 0,075d d

m 0,207n 0,541

⎫= = ⎪⎪

= ⎬⎪= ⎪⎭

→ a 1,5ε ∼ ‰ → 0,22μ =

1 2

u

u

d d 0,100d d

m 0,224n 0,585

⎫= = ⎪⎪

= ⎬⎪= ⎪⎭

→ a 1,5ε ∼ ‰ → 0,26μ =

a a,pret1,5 ‰ε = ε∼

0,22 0,26 0,242+

μ = =

Potrebna površina armature:

2ba,pot

av

f 2,55A b d 0,24 30 80 36,72 cmf 40

= μ ⋅ ⋅ ⋅ = ⋅ ⋅ ⋅ =

2a1,pot a2,.pot a,pot

1 1A A A 36,72 18,36 cm2 2

= = ⋅ = ⋅ =

Usvojena armatura: 6 20± φ ( )2a1 a2A A 18,85 cm= =


PBAB 87 32


1 6O20

2 2O12

3 O8/20

1 6O20

1 2 vil 1d d a 1,0 2,5 0,8 2,0 1,0 6,3 cm pret d 7 cm= = + φ + φ + = + + + = < =

MB 30

RA 400/500

a = 2,5 cm


PBAB 87 33

Zadatak 11.

Dimenzionirati kratki centrično pritisnuti stup.

Statički utjecaji:

g

p

N 600 kNtlak

N 300 kN

= − ⎫⎪ −⎬= − ⎪⎭


Materijal:

MB 40

RA 400/500

_______________________________________________________________

Traži se: Aa,pot _______________________________________________________________

a 0ε < ‰ → g p1,9 ; 2,1γ = γ =

Ultimna tlačna sila:

u g g p pN N N 1,9 600 2,1 300 1770 kN= γ ⋅ + γ ⋅ = ⋅ + ⋅ =

Tlačna sila koju može preuzeti betonski presjek u stanju granične nosivosti:

bu b bN A f 30 30 2,55 2295 kN= ⋅ = ⋅ ⋅ =

u buN N< → minimalno armiranje presjeka min 0,3%μ = ( )bu bfσ <

Minimalna površina uzdužne armature:

a,min 2min a,min min b

b

A 0,3A A 30 30 2,7 cmA 100

μ = → = μ ⋅ = ⋅ ⋅ =

Usvojena armatura:

4 12φ ( )2aA 4,52 cm=


1 4O12

2 O8/20

MB 30

RA 400/500

a = 2,5 cm


PBAB 87 34

Zadatak 12.

Dimenzionirati kratki centrično pritisnuti stup.

Statički utjecaji:

g

p

N 600 kNtlak

N 300 kN

=− ⎫⎪ −⎬= − ⎪⎭


Materijal:

MB 30

RA 400/500 _______________________________________________________________

Traži se: Aa,pot _______________________________________________________________

a 0ε < ‰ → g p1,9 ; 2,1γ = γ =

Ultimna tlačna sila:

u g g p pN N N 1,9 600 2,1 300 1770 kN= γ ⋅ + γ ⋅ = ⋅ + ⋅ =

Tlačna sila koju može preuzeti betonski presjek u stanju granične nosivosti:

bu b bN A f 25 25 2,05 1281,25 kN= ⋅ = ⋅ ⋅ =

>u buN N

Tlačna sila koju u stanju granične nosivosti treba preuzeti uzdužna armatura:

au u buN N N 1770 1281,25 488,75 kN= − = − =

Potrebna površina uzdužne armature:

2aua,pot

av

N 488,75A 12,22 cmf 40

= = =

Usvojena armatura:

4 20φ ( )2aA 12,57 cm=

Geometrijski stupanj armiranja presjeka:

a0 min

b

A 12,57 100 2,01% 0,6 %A 25 25

μ = = ⋅ = > μ =⋅

( )bu bfσ ≥


1 4O20

2 O8/20

MB 30

RA 400/500

a = 2,5 cm


PBAB 87 35

Zadatak 13.


Statički utjecaji:

g

p

g p

M 120 kNm

M 100 kNm

N N 0

=

=

= =

Presjek: pravokutni b/d = 30/d cm

Materijal:

MB 30

RA 400/500 _______________________________________________________________

Traži se:

h (d) i

a,potA _______________________________________________________________

b bu

a

3,5 ‰najmanja statička visina

3 ‰

ε = ε = ⎫⎪→⎬ε = ⎪⎭

a 3ε = ‰ → g p1,6 ; 1,8γ = γ =

u g g p pM M M 1,6 120 1,8 100 372 kNm= γ ⋅ + γ ⋅ = ⋅ + ⋅ =


( )au u uM M 372 kNm N 0= = =


au au

z z

m m 0,338

k k 0,776

∗

∗

= =

= = - granica 1-struko armiranih presjeka

au auau pot2

au bb

M M 372m h 42,3 cmm b f 0,338 0,3 2,05b h f

= → = = =⋅ ⋅ ⋅ ⋅⋅ ⋅

Potrebna površina vlačne armature: 2

2aua,pot

z av

M 372 10A 28,3 cmk h f 0,776 42,3 40

⋅= = =

⋅ ⋅ ⋅ ⋅

Usvojena armatura:

6 25φ ( )2aA 29,45 cm=

2 vild a 1 2,5 0,8 2,5 1 6,8 cm= + φ + φ + = + + + =


PBAB 87 36

pot pot 2d h d 42,3 6,8 49,1cm= + = + =

Usvojeno: d = 50 cm


1 6O25

2 2O12

3 O8/20

Dimenzioniranje bez uporabe tabela

b bu

a

3,5 ‰najmanja statička visina

3 ‰

ε = ε = ⎫⎪→⎬ε = ⎪⎭


ravnoteže: * *

au bu* *bu B x b*

z

2 auau x z b pot

x z b

M P z

P b x f b k h f

z k h

MM k k b h f hk k b f

∗

∗

∗

∗∗ ∗ ∗

∗ ∗

= ⋅

= α ⋅ ⋅ ⋅ = α ⋅ ⋅ ⋅ ⋅

= ⋅

= α ⋅ ⋅ ⋅ ⋅ ⋅ → =α ⋅ ⋅ ⋅ ⋅


b 2ε ≥ ‰ → b

b

3 2 3 3,5 2 0,80953 3 3,5⋅ ε − ⋅ −

α = = =⋅ ε ⋅


bx

b a

3,5k 0,5383,5 3,0

∗ ε= = =ε + ε +


z p xk 1 k k 1 0,416 0,538 0,776∗ ∗ ∗= − ⋅ = − ⋅ =

b 2ε ≥ ‰ → ( ) ( )

2 2b b

pb b

3 4 2 3 3,5 4 3,5 2k 0,4162 3 2 2 3,5 3 3,5 2

∗ ⋅ ε − ⋅ ε + ⋅ − ⋅ += = =

⋅ ε ⋅ ⋅ ε − ⋅ ⋅ ⋅ −

b = 0,30 m

MB 30

RA 400/500

a = 2,5 cm


PBAB 87 37

b 2 2


= =

pot372h 42,3 cm

0,8095 0,538 0,776 0,3 2,05= =

⋅ ⋅ ⋅ ⋅

Potrebna površina vlačne armature: 2

2aua,pot

z av

M 372 10A 28,3 cmk h f 0,776 42,3 40

⋅= = =

⋅ ⋅ ⋅ ⋅

Usvojena armatura:

6 25φ ( )2aA 29,45 cm=

2 vild a 1 2,5 0,8 2,5 1 6,8 cm= + φ + φ + = + + + =

pot pot 2d h d 42,3 6,8 49,1cm= + = + =

Usvojeno: d = 50 cm


PBAB 87 38

Zadatak 14.

Sračunati moment koji presjek može preuzeti u stanju granične otpornosti na osnovu

zadanih podataka.

1 5O19

2 2O12

3 O8/20

_______________________________________________________________

Traži se: Mu i


Postupak pomoću ''mau'' tabela

Udaljenost težišta vlačne armature do vlačnog ruba:

2 vil1,9d a 2,5 0,8 4,25cm

2 2φ

= + φ + = + + =

Statička visina presjeka:

2h d d 50 4,25 45,75cm= − = − =

b a ava Ma Ma au

av b

f A f 14,18 40A b h 0,202 m 0,1808f b h f 30 45,75 2,05

= ω ⋅ ⋅ ⋅ → ω = ⋅ = ⋅ = → =⋅ ⋅

2 2auau au au b2

b

Mm M m b h f 0,1808 0,3 45,75 2,05 232,7 kNmb h f

= → = ⋅ ⋅ ⋅ = ⋅ ⋅ ⋅ =⋅ ⋅

Postupak bez uporabe tabela

Pbu

Zau

Mu

Tb

Pretpostavka: a av au avfε > ε → σ =

a av x bA f b k h f 0⋅ − α ⋅ ⋅ ⋅ ⋅ =

MB 30

RA 400/500

a = 2,5 cm

au bu

au au a

bu b x b

a au x b

H 0

Z P 0

Z A

P b x f b k h f

A b k h f 0

=

− =

= σ ⋅

= α ⋅ ⋅ ⋅ = α ⋅ ⋅ ⋅ ⋅

⋅ σ − α ⋅ ⋅ ⋅ ⋅ =

∑


PBAB 87 39

2 vil1,9d a 2,5 0,8 4,25 cm

2 2φ

= + φ + = + + =

2h d d 50 4,25 45,75 cm= − = − =

Pretpostavka: bb

b

3 22‰ 3,5 ‰3⋅ ε −

< ε ≤ → α =⋅ ε

bx

a b

k ε=ε + ε

b ba av b

b a b

3 2A f b h f 03⋅ ε − ε

⋅ − ⋅ ⋅ ⋅ ⋅ =⋅ ε ε + ε

( )2aA 14,18 cm 5 19= φ

b 30 cm=

( )( )

2b

2av

f 2,05 kN cm MB30

f 40 kN cm RA 400 / 500

=

=

( )

( ) ( )

b

a b

ba b

a b

a b b

a b

a b

3 214,18 40 30 45,75 2,05 03

3 2567,2 937,875 0

567,2 567,2 2813,625 1875,75 0

567,2 2246,425 1875,75 0 : 567,2

3,9606 3,307 0

⋅ ε −⋅ − ⋅ ⋅ ⋅ =

⋅ ε + ε

⋅ ε −− ⋅ = ⋅ ε + ε

ε + ε

⋅ ε + ⋅ ε − ⋅ ε + =

⋅ ε − ⋅ ε + =

ε − ⋅ ε + =

Pretpostavka: b bu 3,5 ‰ε = ε =

a 3,9606 3,5 3,307 0ε − ⋅ + = →

a au10,56 ‰ 10 ‰ε = > ε =

Pretpostavka: a au 10 ‰ε = ε =

b

bub

10 3,9606 3,307 0

3,5 ‰3,36 ‰

2‰

− ⋅ ε + = →

≤ ε =⎧⎪ε = ⎨>⎪⎩

2 u buM 0 M P z= → = ⋅∑

bu b x bP b x f b k h f= α ⋅ ⋅ ⋅ = α ⋅ ⋅ ⋅ ⋅

b

b

3 2 3 3,36 2 0,80163 3 3,36⋅ ε − ⋅ −

α = = =⋅ ε ⋅

bx

a b

3,36k 0,251510 3,36

ε= = =ε + ε +

zz k h 0,8961 45,75 41cm= ⋅ = ⋅ = 2 2b b

p 2 2b b

3 4 2 3 3,36 4 3,36 2k 0,41316 4 6 3,36 4 3,36⋅ ε − ⋅ ε + ⋅ − ⋅ +

= = =⋅ ε − ⋅ ε ⋅ − ⋅


PBAB 87 40

z p xk 1 k k 1 0,4131 0,2515 0,8961= − ⋅ = − ⋅ =

buP 30 0,2515 45,75 0,8016 2,05 567,2 kN= ⋅ ⋅ ⋅ ⋅ =

uM 567,2 0,41 232,55 kNm= ⋅ =

Ili:

1 au uM 0 Z z M= → ⋅ =∑

au a avZ A f 14,18 40 567,2 kN= ⋅ = ⋅ =

zz k h 0,896 45,75 41cm= ⋅ = ⋅ =

uM 567,2 0,41 232,55 kNm= ⋅ =

Kontrola:

au bu au buH 0 Z P 0 Z P 567,2 567,2= → − = → = → =∑


PBAB 87 41

Zadatak 15.

Sračunati (bez uporabe tabela) presječne sile (Mu, Nu) koje presjek može preuzeti u

stanju granične otpornosti na osnovu zadanih podataka.

21

A'a

εb

εa'

εa

σbu

σau

σau'

Aa

____________________________________________

Traži se: Mu i Nu Pbu

Zau

Mu

Tb

12

N (vlak)u

Pau

1

au a avZ A f 19,01 40 760,4 kN= ⋅ = ⋅ =

bb

b

3 2 3 2,8 22‰ 0,7623 3 2,8⋅ ε − ⋅ −

ε > → α = = =⋅ ε ⋅

bx

a b

2,8k 0,21910 2,8

ε= = =ε + ε +

bu x bP k b h f 0,762 0,219 30 45 2,05 461,8 kN= α ⋅ ⋅ ⋅ ⋅ = ⋅ ⋅ ⋅ ⋅ =

2h d d 50 5 45 cm= − = − =

'b a

1x x dε ε

= →−

' 1a b

x d 9,85 5 2,8 1,379 ‰x 9,85− −

ε = ⋅ ε = ⋅ =

xx k h 0,219 45 9,85 cm= ⋅ = ⋅ =

' 'au a a

1,379E 210000 289,6 MPa1000

σ = ε ⋅ = ⋅ =

' 'au a auP A 6,03 28,96 174,6 kN= ⋅ σ = ⋅ =

MB 30

RA 400/500

b/d = 30/50 cm

d1 = 5,0 cm

d2 = 5,0 cm

εa = εau = 10 ‰

εb = 2,8 ‰

Aa = 19,01 cm2

Aa' = 6,03 cm2

( )

au bu au u

u au bu au

au a au a av a av

bu b x b

' 'au a au

H 0

Z P P N 0

N Z P P

Z A A f

P b x f b k h f

P A

= →

− − − =

= − −

= ⋅ σ = ⋅ ε > ε

= α ⋅ ⋅ ⋅ = α ⋅ ⋅ ⋅ ⋅

= ⋅ σ

∑


PBAB 87 42

u au bu auN Z P P 760,4 461,8 174,6 124,0 kN= − − = − − = - dobro pretpostavljen predznak sile.

( )( )

1

u au 1 bu u a

u au 1 bu u a

M 0

M P h d P z N y 0

M P h d P z N y

= →

− ⋅ − − ⋅ − ⋅ = →

= ⋅ − + ⋅ + ⋅

∑

b 2‰ε > →

2 2b b

p 2 2b b

3 4 2 3 2,8 4 2,8 2k 0,46 4 6 2,8 4 2,8⋅ ε − ⋅ ε + ⋅ − ⋅ +

= = =⋅ ε − ⋅ ε ⋅ − ⋅

z p xk 1 k k 1 0,4 0,219 0,912= − ⋅ = − ⋅ =

zz k h 0,912 45 41,04 cm= ⋅ = ⋅ =

( )uM 174,6 0,45 0,05 461,8 0,4104 124,0 0,20 284,2 kNm= ⋅ − + ⋅ + ⋅ =

u

u

N 124,0 kN vlak

M 284,2 kNm

= −

=


PBAB 87 43

Zadatak 16.


Statički utjecaji:

g

p

g p

M 500 kNm

M 300 kNm

N N 0

=

=

= =

Materijal:

MB 30

RA 400/500 _____________________________________________________________________

Traži se:

a,potA i

b a au( 10ε ε = ε = ‰) ili a b bu( 3,5ε ε = ε = ‰) _____________________________________________________________________


u g g p pM M M 1,6 480 1,8 320 1344 kNm= γ ⋅ + γ ⋅ = ⋅ + ⋅ =


( )au u uM M 1344 kNm N 0= = =




Pretpostavimo da neutralna os pada u ploču ''T'' presjeka (x < dpl), tj. da je pritisnuta

zona betona pravokutnog oblika širine b = 90 cm: 2

auau au2 2

b

M 1344 10m 0,153 m 0,338b h f 90 69 2,05

∗⋅= = = < =


' 'T'' – presjek:

b = 90 cm

bo = 30 cm

d = 80 cm

dpl = 12 cm


PBAB 87 44

mau = 0,153 → kx = 0,22

Kontrola položaja neutralne osi: '

x plx k h 0,220 69 15,2 cm d 12 cm ''T '' presjek= ⋅ = ⋅ = > = → −

Dimenzioniranje presjeka ''T'' oblika tlačne betonske zone

0

b 90 3,0 5b 30

= = < → postupak reducirane širine

pl

x,pret b

0

d 12 0,174h 69

k 0,220 0,95

b 90 3,0b 30

⎫= = ⎪

⎪⎪= → λ =⎬⎪⎪= = ⎪⎭

- faktor redukcije

Širina zamjenjujućeg pravokutnog presjeka:

i bb b 0,95 90 85,5 cm= λ ⋅ = ⋅ =

2au

au au2 2i b

M 1344 10m 0,161 m 0,338b h f 85,5 69 2,05

∗⋅= = = < =

⋅ ⋅ ⋅ ⋅- 1-struko armiran presjek

( )au x x,pret pretm 0,161 k 0,229 k 0,220 h h= → = > = <

x,pret b

0

d 12 0,174h 69k 0,25 0,89

b 90 3,0b 30

⎫= = ⎪

⎪⎪= → λ =⎬⎪⎪= = ⎪⎭

i bb b 0,89 90 80,1cm= λ ⋅ = ⋅ =

2au

au au2 2i b

M 1344 10m 0,172 m 0,338b h f 80,1 69 2,05

∗⋅= = = < =

⋅ ⋅ ⋅ ⋅- 1-struko armiran presjek

( )x x,pret pret

Maau

a

b

k 0,241 k 0,250 h h

0,191m 0,172

10 ‰

3,18 ‰

⎧ = < = >⎪⎪ω =⎪= → ⎨ε =⎪⎪ε =⎪⎩


2ba,pot Ma

av

f 2,05A b h 0,191 80,1 69 54,1cmf 40

= ω ⋅ ⋅ ⋅ = ⋅ ⋅ ⋅ =

Usvojena armatura:

12 25φ ( )2aA 58,92 cm=


PBAB 87 45


1 12O25

2 2O12

3 O10/20

2 2O12

2 vil

2 pret

2,5d a 3,0 2,5 1,0 2,5 3,0 10,25 cm2 2

h d d 80 10,25 69,75 cm h 11cm

φ= + φ + φ + + = + + + + =

= − = − = > =

MB 30

RA 4000/500

a = 2,5 cm


PBAB 87 46

Zadatak 17.


Statički utjecaji:

g

p

g p

M 600 kNm

M 500 kNm

N N 0

=

=

= =

Presjek: 0

pl

b 200 cm

b 30 cm''T ''

d 10 cm

d 80 cm

=⎧⎪

=⎪⎨

=⎪⎪

=⎩

Materijal:

MB 30

RA 400/500 _____________________________________________________________________

Traži se:

a,potA i

b a au( 10ε ε = ε = ‰) ili a b bu( 3,5ε ε = ε = ‰) _____________________________________________________________________


u g g p pM M M 1,6 600 1,8 500 1860 kNm= γ ⋅ + γ ⋅ = ⋅ + ⋅ =


( )au u uM M 1860 kNm N 0= = =



pret 2h d d 80 11 69cm= − = − = - pretpostavljena statička visina presjeka

Pretpostavimo da neutralna os pada u ploču ''T'' presjeka (x < dpl), tj. da je pritisnuta


auau au2 2

b

M 1860 10m 0,095 m 0,338b h f 200 69 2,05

∗⋅= = = < =


mau = 0,095 → kx = 0,157


xx k h 0,157 69 10,8 cm d 10 cm ''T '' presjek= ⋅ = ⋅ = > = → −


PBAB 87 47

Dimenzioniranje presjeka ''T'' oblika tlačne betonske zone

0

b 200 6,7 5b 30

= = > → postupak dimenzioniranja ''T''- presjeka uz zanemarenje

napona tlaka u betonu rebra presjeka

Krak unutrašnjih sila:

pld 10z h 69 64 cm2 2

= − = − =


2aua,pot

av

M 1860A 72,66 cmz f 0,64 40

= = =⋅ ⋅

Usvojena armatura:

12 28φ ( )2aA 73,89 cm=


1 12O28

2 2O12

3 O10/20

2 2O12

2 vil

2 pret

2,8d a 3,0 2,5 1,0 2,8 3,0 10,7 cm2 2

h d d 80 10,7 69,3 cm h 69 cm

φ= + φ + φ + + = + + + + =

= − = − = < =

Kontrola tlačne betonske zone

bu b

aubu

2b pl

bubu b2 2

b

f

M 1860P 2906,25 kNz 0,64

A b d 200 10 2000 cm

P 2906,25 kN kN1,45 f 2,05A 2000 cm cm

σ ≤

= = =

= ⋅ = ⋅ =

σ = = = < =

MB 30

RA 400/500

a = 2,5 cm


PBAB 87 48

Zadatak 18.


Statički utjecaji:

g

p

g

p

M 1600 kNm

M 1000 kNm

N 600 kNtlak

N 400 kN

=

=

= − ⎫⎪ −⎬= − ⎪⎭

Presjek: 0

pl

b 250 cm

b 40 cm''T ''

d 12 cm

d 120 cm

=⎧⎪

=⎪⎨

=⎪⎪

=⎩

Materijal:

MB 30

RA 400/500 _______________________________________________________________

Traži se:

a,potA i



u g g p pM M M 1,6 1600 1,8 1000 4360 kNm= γ ⋅ + γ ⋅ = ⋅ + ⋅ =

( ) ( )u g g p pN N N 1,6 600 1,8 400 1680 kN= γ ⋅ + γ ⋅ = ⋅ − + ⋅ − = −


au u u aM M N y= − ⋅

Tb

22

a

T


PBAB 87 49

Pretpostavka: d2 = 11 cm

i iT

i

A a 250 12 114 108 40 54y 78,6 cmA 250 12 108 40⋅ ⋅ ⋅ + ⋅ ⋅

= = =⋅ + ⋅

∑∑

a T 2,prety y d 78,6 11 67,6 cm= − = − =

( )auM 4360 1680 0,676 5495,7 kNm= − − ⋅ =


pret 2h d d 120 11 109cm= − = − =

Pretpostavimo da neutralna os pada u ploču ''T'' presjeka (x < d), tj. da je pritisnuta


auau au2 2

b

M 5495,7 10m 0,090 m 0,338b h f 250 109 2,05

∗⋅= = = < =


mau = 0,090 → kx = 0,152


xx k h 0,152 109 16,6 cm d 12 cm ''T '' presjek= ⋅ = ⋅ = > = → −

0

b 250 6,25 5b 40

= = > → postupak dimenzioniranja ''T''- presjeka uz zanemarenje

napona tlaka u betonu rebra presjeka

Krak unutrašnjih sila:

d 12z h 109 103 cm2 2

= − = − =


2au ua,pot

av av

M N 5495,7 1680A 133,39 42 91,39 cmz f f 1,03 40 40

−= + = + = − =

⋅ ⋅

Usvojena armatura:

15 28φ ( )2aA 92,36 cm=

Kontrola tlačne betonske zone

bu b

aubu

2b pl

bubu b2 2

b

f

M 5495,7P 5335,6 kNz 1,03

A b d 250 12 3000 cm

P 5335,6 kN kN1,78 f 2,05A 3000 cm cm

σ ≤

= = =

= ⋅ = ⋅ =

σ = = = < =


PBAB 87 50


1 15O28

3 2O12

2 O10/20

3 2O12

3 2O12

2 vil

2 pret

2,8d a 3,0 2,5 1,0 2,8 3,0 10,7 cm2 2

h d d 120 10,7 109,3 cm h 109 cm

φ= + φ + φ + + = + + + + =

= − = − = > =

MB 30

RA 400/500

a = 2,5 cm


PBAB 87 51

Zadatak 19.

Sračunati moment koji može preuzeti 1-struko armirani AB presjek i potrebnu količinu

armature na osnovu zadanih podataka.

Presjek: 0

pl

b 120 cm

b 30 cm

''T '' d 10 cm

d 70 cm

h 60 cm

=⎧⎪

=⎪⎪

=⎨⎪

=⎪⎪

=⎩

Materijal:

MB 30

RA 400/500 _____________________________________________________________________

Traži se: ∗auM i a,potA∗

_____________________________________________________________________ b bu

a

3,5 ‰

3 ‰

ε = ε =

ε =

bx

a b

3,5k 0,5383,0 3,5

∗ ε= = =

ε + ε +

xx k h 0,538 60 32,3 cm d 10 cm∗ ∗= ⋅ = ⋅ = > = → ''T''- presjek

A) Rastavljanje ''T''- presjeka na pravokutne presjeke

A.1

0

0

Presjek 1 Presjek 2

pl

0 0

pl

pl

Presjek 1

0

εb

εa

fbu

Pbu(1)

Zau(1)

1


PBAB 87 52

b buau x z

a

MB30

3,5 ‰m 1,20 ; k 0,538 ; k 0,776

3 ‰∗ ∗ ∗

ε = ε = ⎫⎪ → = = =⎬ε = ⎪⎭

* 2 2au(1) au bM m b h f 0,338 1,2 65 2,05 3513 kNm∗ = ⋅ ⋅ ⋅ = ⋅ ⋅ ⋅ =

1 zz k h 0,776 65 50,44 cm∗ ∗= ⋅ = ⋅ =

au(1)bu(1)

1

M 3513P 6964,7 kN0,5044z

∗∗

∗= = =

Presjek 2

εb1

εa

Pbu(2)

Zau(2)

1

0

0

σbuεb2

fav

2

plb2 b1 bu b

x d 35 103,5 2,5 ‰ f35x

∗

∗

− −ε = ε ⋅ = ⋅ = → σ =

( ) ( )bu(2) 0 pl 2 bP b b x d f∗ ∗= − ⋅ − ⋅ α ⋅

b2b2 2

b2

3 2 3 2,5 22‰ 0,7333 3 2,5⋅ ε − ⋅ −

ε > → α = = =⋅ ε ⋅

( ) ( )bu(2)P 120 30 35 10 0,733 2,05 3381kN∗ = − ⋅ − ⋅ ⋅ =

2 pl 2z h d a 65 10 9,77 45,2 cm∗ ∗= − − = − − =

( ) ( )2 p2 pla k x d 0,391 35 10 9,77 cm∗ ∗ ∗= ⋅ − = ⋅ − =

b22 2b2 b2

p2 2 2b2 b2

2‰

3 4 2 3 2,5 4 2,5 2k 0,3916 4 6 2,5 4 2,5

∗

ε > →

⋅ ε − ⋅ ε + ⋅ − ⋅ += = =

⋅ ε − ⋅ ε ⋅ − ⋅

au(2) bu(2) 2M P z 3380,96 0,452 1528,2 kNm∗ ∗ ∗= ⋅ = ⋅ =

Vrijednosti ultimnih djelovanja za ''T''- presjek:

bu(T) bu(1) bu(2)P P P 6964,7 3381 3583,7 kN∗ ∗ ∗= − = − =

au(T) au(1) au(2)M M M 3513 1528,20 1984,8 kN∗ ∗ ∗= − = − =

au(T)T

bu(T)

M 1984,8z 0,554 m3583,7P

∗∗

∗= = =


PBAB 87 53


au(T) 2a,pot

T av

M 1984,8A 89,6 cm0,554 40z f

∗

∗= = =⋅⋅

A.2

0

pl

0

0Presjek 1Presjek 2

0

pl

Presjek 1

εb

εa

fbu

P*bu(1)

Z*au(1)

1

0

b buau x z

a

MB30

3,5 ‰m 0,338 ; k 0,538 ; k 0,776

3 ‰∗ ∗ ∗

ε = ε = ⎫⎪ → = = =⎬ε = ⎪⎭

2* 2 2

au(1) au 0 b 2

0,3 65M m b h f 0,338 0,3 65 2,05 878,25 kNm1,2

∗ ⋅= ⋅ ⋅ ⋅ = ⋅ ⋅ ⋅ = =

1 zz k h 0,776 65 50,44 cm∗ ∗= ⋅ = ⋅ =

au(1)bu(1)

1

M 878,25P 1741,2 kN0,5044z

∗∗

∗= = =

Presjek 2

εb1

εa

P*bu(2)

Z*au(2)

2

σ =bu

εb2

fav

fb

σ =au

0

pl

0


PBAB 87 54

plb2 b1 bu b

x d 35 103,5 2,5 ‰ f35x

∗

∗

− −ε = ε ⋅ = ⋅ = → σ =

( ) ( )bu(2) 0 pl bP b b d f 120 30 10 2,05 1845 kN∗ = − ⋅ ⋅ = − ⋅ ⋅ =

pl2

d 10z h 65 60 cm2 2

∗ = − = − =

au(2) bu(2) 2M P z 1845 0,6 1107 kNm∗ ∗ ∗= ⋅ = ⋅ =

Vrijednosti ultimnih djelovanja za ''T''- presjek:

bu(T) bu(1) bu(2)P P P 1741,2 1845 3586,2 kN∗ ∗ ∗= + = + =

au(T) au(1) au(2)M M M 878,25 1107 1985,25 kN∗ ∗ ∗= + = + =

au(T)T

bu(T)

M 1985,25z 0,554 m3586,2P

∗∗

∗= = =


au(T) 2a,pot

T av

M 1985,25A 89,6 cm0,554 40z f

∗

∗= = =⋅⋅

B) Rješenje zadatka pomoću jednadžbi za proračun ''T''- presjeka

Moment savijanja kojeg može preuzeti ''T''- presjek 1-struko armiran:

2au(T) b x z(T)M b h f k k∗ ∗ ∗ ∗= ⋅ ⋅ ⋅ ⋅ ⋅ λ

b1

b21

b2

2 2b2 b2

p1 2 2b2 b2

3,5 ‰

3 2 3 3,5 2 0,80953 3 3,5

3 4 2 3 3,5 4 3,5 2k 0,4166 4 6 3,5 4 3,5

∗

ε = →

⋅ ε − ⋅ −α = = =

⋅ ε ⋅

⋅ ε − ⋅ ε + ⋅ − ⋅ += = =

⋅ ε − ⋅ ε ⋅ − ⋅

b1

b22

b2

2 2b2 b2

p2 2 2b2 b2

2,5 ‰

3 2 3 2,5 2 0,7333 3 2,5

3 4 2 3 2,5 4 2,5 2k 0,3916 4 6 2,5 4 2,5

∗

ε = →

⋅ ε − ⋅ −α = = =

⋅ ε ⋅

⋅ ε − ⋅ ε + ⋅ − ⋅ += = =

⋅ ε − ⋅ ε ⋅ − ⋅

p(T)k∗

∗∗

ϕ=

λ


PBAB 87 55

pl pl pl01 p1 2 p2

x x x

d d dbk 1 1 k 1

b k h k h k h

30 100,8095 0,416 0,733 1 1120 0,538 65

10 100,391 1 0,1150,538 65 0,538 65

∗ ∗ ∗∗ ∗ ∗

⎡ ⎤⎛ ⎞ ⎛ ⎞⎛ ⎞ϕ = α ⋅ − α ⋅ − ⋅ − ⋅ + ⋅ − =⎢ ⎥⎜ ⎟ ⎜ ⎟⎜ ⎟ ⋅ ⋅ ⋅⎝ ⎠ ⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

⎛ ⎞⎛ ⎞= ⋅ − ⋅ − ⋅ − ⋅⎜ ⎟ ⎜ ⎟⋅⎝ ⎠ ⎝ ⎠

⎡ ⎤⎛ ⎞⋅ + ⋅ − =⎢ ⎥⎜ ⎟⋅ ⋅⎝ ⎠⎣ ⎦

pl01 2

x

db 30 101 1 0,8095 0,733 1 1 0,417b 120 0,538 65k h

∗∗

⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞λ = α − α ⋅ − ⋅ − = − ⋅ − ⋅ − =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⋅⋅ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠

p(T)

z(T) p(T) x

0,115k 0,27580,417

k 1 k k 1 0,2758 0,538 0,8516

∗∗

∗

∗ ∗ ∗

ϕ= = =

λ

= − ⋅ = − ⋅ =

2 2au(T) b x z(T)M b h f k k 1,2 65 2,05 0,538 0,8516 0,417 1985,7 kNm∗ ∗ ∗ ∗= ⋅ ⋅ ⋅ ⋅ ⋅ λ = ⋅ ⋅ ⋅ ⋅ ⋅ =


2au(T) au(T) 2

a,pot(T) av z(T) av

M M 1985,7 10A 89,68 cm0,8516 65 40z f k h f

∗ ∗

∗ ∗

⋅= = = =

⋅ ⋅⋅ ⋅ ⋅

dimenzioniranje presjeka pbab-87 zad 1-19

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