Betonarme 1 Ders2 Ing

REINFORCED CONCRETE 10423522

Dr. Murat Serdar Kırçıl

İNM 1-024

www.yildiz.edu.tr/~kircil

Betonarme 1 Ders Notları

Yrd.Doç.Dr.Murat Serdar Kırçıl

Fundamental principles and methods for design of reinforced concrete structures

Elastic design (working stress design / Allowable stress design)

The fundamentals of this method has been established in the beginning of

20th century. Both steel and concrete are assumed to be linear elastic and Hooke Law is valid.

This assumption may be valid for service loads (işletme yükleri)

However; it’s not valid anymore in case of failure. For instance, under

the effect of a strong ground motion, the assumption of material

linearity is not valid.

σ

ε

σ = EεE

Analysis and design of a cross-section



Both steel and concrete are assumed to be linear elastic and Hooke Law is valid.

Main assumptions of this method

Concrete can not take any tension (REALISTIC)

Stresses in steel and concrete is considerably changed with time due

to time dependent deformations. That’s why, stresses can not be

calculated with reasonable accuracy.

The little tensile strength of concrete is neglected. However, in the

design of some structures like water tanks, tensile stress is limited for

leak proofing.



Strength of materials are decreased by safety factors to obtain allowable stresses. The structural element is assumed to be safe if

the maximum stress of section is lower than the allowable stress (NOT

REALISTIC)



Bernouilli-Navier hypothesis is valid (REALISTIC)

Plane sections remain plane after bending.






This assumption is valid for service loads. In case of failure it doesn't

represent the real behavior of the RC sections.

fc

εεεεcoεεεεcu

Stress (σ)

Deformation(ε)Valid for this region

Both steel and concrete are assumed to be linear elastic and Hooke Law is valid (NOT REALISTIC)






It’s not a realistic assumption since concrete has not a constant elastic

modulus.

There is a modular ratio between concrete and steel reinforcement (NOT REALISTIC!)






Limit state design (Limit durumlara göre hesap)

Limit state defines a particular state of a structure which may be

collapse or loss of function, either partial or total. Structures must

be designed so that the probability of reaching a limit state is lower

than prescribed level.

ULTIMATE LIMIT STATE (Son limit durum)

The collapse of a part of a structure or the whole structure.

SERVICEABILITY LIMIT STATE (Kullanılabilirlik limit durumu)

The loss of a function of a structure because of excessive deformation,

wide cracks or vibration under the effect of service loads (dead+live).

For instance, a slab which doesn’t collapse but vibrates.


In limit state design, the probability of reaching a limit state is restricted

with a prescribed level (%5-%10) for the design of structures.

The first stage of design making design of a cross-section for the

ultimate limit state and the following step is checking the serviceability

requirements (deformations and cracks).



Cross-sectional design based on the limit state design assumptions is

called Ultimate Strength Theory or Ultimate Strength Design.

Concrete does not take any tension, all tensile stresses are taken by reinforcement.

Bernouilli-Navier hypothesis is valid.

Concrete σ-ε curve is the curve obtained from uniaxialcompression. Reinforcement steel is elastoplastic.

fc

εεεεcoεεεεcu

Gerilme (σ)

Şekil değiştirme (ε)

fy

εεεεsy εεεεsu

Gerilme (σ)

Şekil değiştirme (ε)

Loads are increased by load factors and material strength isdecreased by material factors. Load and material factors arecalled partial safety factors.






Probability of reaching or exceeding a limit state of structural

system elements is restricted in limit state design.

What is the

probability of

reaching a limit

state?

Probability of reaching a limit state is 10-6

The safety of structure can be expressed as given below (R and F are

strength and load, respectively)

FR≥ structure is assumed to be safeIf

610)FR(P

−=<Probability of reaching a limit state is





Why do we need a probabilistic

approach?

Load and strength are probabilistic variables.

Probability of reaching or exceeding a limit state of structural

system elements is restricted in limit state design.





We need to know probability distribution of load and strength aswe can determine the safety of structures by using a probabilisticapproach. The probability distribution of them is usuallyassumed as normal (gauss distribution).

Load and strength are probabilistic variables.

Why do we need a probabilistic

approach?





0 100 200 300 400 500 600 700 800 900 1000 1100 1200 1300

Değişken

0

1

2

3

4

5

6

7

Gözle

m s

ayıs

ı

( )

−−

=

2mx

2

1

x e2

1xf

σ

πσ

m mean value, σ standard deviation

We need to know probability distribution of load and strength aswe can determine the safety of structures by using a probabilisticapproach. The probability distribution of them is usuallyassumed as normal (gauss distribution).





( )

−−

=

2mx

2

1

x e2

1xf

σ

πσ

smx

=−

σ

( )

−

=

2s

2

1

x e2

1xf

π Toplam alan=1

0 100 200 300 400 500 600 700 800 900 1000 1100 1200 13000

1

2

3

4

5

6

7

0

Standard normal

distribution m=0, σ =1

s

0 100 200 300 400 500 600 700 800 900 1000 1100 1200 13000

1

2

3

4

5

6

7

mx





smx

=−

σ( )

−

=

2s

2

1

x e2

1xf

π

0 100 200 300 400 500 600 700 800 900 1000 1100 1200 13000

1

2

3

4

5

6

7

0

AsmA

=−

σBs

mB=

−

σ

sA sB

( )BxAp <<

CsmC

=−

σ

sC

( )xCp <

Standard normal

distribution m=0, σ =1

Probability of having a value

between A and B for x?

Probability of having a value

higher than C for x?





smx

=−

σ( )

−

=

2s2

1

x e2

1xf

π

smx σ+=

0 100 200 300 400 500 600 700 800 900 1000 1100 1200 13000

1

2

3

4

5

6

7

0-2 -1-3-4 4321s-2σ -1σ-3σ 4σ3σ2σ1σ-4σ

sσ





smx

=−

σ

smx σ+=

0 100 200 300 400 500 600 700 800 900 1000 1100 1200 13000

1

2

3

4

5

6

7

0-2 -1-3-4 4321s-2σ -1σ-3σ 4σ3σ2σ1σ-4σ

sσ

0 100 200 300 400 500 600 700 800 900 1000 1100 1200 13000

1

2

3

4

5

6

7

FmFk

sFF mk σ+=

Characteristic Load: The load

value with the %10 probability of

being exceeded.

σ+= 28.1FF mk

0.1

( ) 1.0xFp k =<





smx

=−

σ

smx σ+=

0 100 200 300 400 500 600 700 800 900 1000 1100 1200 13000

1

2

3

4

5

6

7

0-2 -1-3-4 4321s-2σ -1σ-3σ 4σ3σ2σ1σ-4σ

sσ

0 100 200 300 400 500 600 700 800 900 1000 1100 1200 13000

1

2

3

4

5

6

7

Rm

sRR mk σ−=

Characteristic strength: Characteristic strength is the

strength below which has 10%

probability will fall.

( ) 1.0Rxp k =<

Rk

σ−= 28.1RR mk

0.1





smx

=−

σ

smx σ+=

0 100 200 300 400 500 600 700 800 900 1000 1100 1200 13000

1

2

3

4

5

6

7

0-2 -1-3-4 4321s-2σ -1σ-3σ 4σ3σ2σ1σ-4σ

sσ

0 100 200 300 400 500 600 700 800 900 1000 1100 1200 13000

1

2

3

4

5

6

7

Fm-2 -1-3-4 Fk





Fm Fk RmRk

The dashed area shows the probability of collapse.

Fk should be increased and Rk should be decreased to reduce the

probability of collapse.

kf

m

k FR

γ≥γ

γm ve γf partial safety factors (kısmi güvenlik katsayıları)

1m >γ 1f >γ

FR≥ structure is assumed to be safeIf





The probability of collapse is assumed to reduce to predetermined value

(10-6) by increasing characteristic loads with load factors (γf) and

decreasing the characteristic strength with material factors (γm).

kf

m

k FR

γ≥γ

1m >γ 1f >γ

γm ve γf partial safety factors (kısmi güvenlik katsayıları)





MALZEME KATSAYILARI – TS500

BETON ÇELİK

5.1mc =γCast in place concrete (yerinde dökme beton)

4.1mc =γ

7.1mc =γ

Precast structural elements (prefabrike elemanlar)

If standard control is not available

15.1ms =γ

mc

ckcd

ff

γ=

ms

yk

yd

ff

γ=

mc

ctkctd

ff

γ=

fck:Characteristic compressive strength of concrete (Betonun karakteristik basınç dayanımı)

fctk: Characteristic tensile strength of concrete (Betonun karakteristik çekme dayanımı)

fyk: Characteristic yield strength of reinforcement steel (Çeliğin karakteristik akma dayanımı)

fcd:Design compressive strength of concrete (betonun hesap basınç dayanımı)

fctd: Design tensile strength of concrete (Betonun hesap çekme dayanımı)

fyd: Design yield strength of reinforcement steel (Çeliğin hesap akma dayanımı)




STRENGTH REDUCTION FACTORS (Kapasite azaltma katsayıları)–ACI318

900 .=φ Pure bending (Basit eğilme)

ACI318 gives strength reduction factors (Ø) instead of material factors

900 .=φ Axial tension (Eksenel çekme)

900700 .. −=φ Axial force and flexure (Bileşik eğilme)

750700 .. −=φ Axial compression (Eksenel basınç)

MATERIAL COEFFICIENTS– CEB

5.1mc =γ

4.1mc =γ

15.1ms =γ




Limit state design (Limit durumlara göre hesap )

LOAD COEFFICIENT – TS500

Q6.1G4.1 +

Only vertical loads

G:Dead load (Dead Load Zati yük / ölü yük))

T2.1Q2.1G0.1 ++ Q:Live load (Hareketli yük)

T:Temparature, differetial settlement, creep (Sıcaklık farkı, Farklı oturma, sünme)

Vetical loads + WindIn addition to (1)

(1)

W3.1Q3.1G0.1 ++

W3.1G9.0 +

W:Wind load (Rüzgar yükü)

In case of earthquake

E0.1Q0.1G0.1 ++

E0.1G9.0 +

E:Deprem yükü (Earthquake Load)

In case of lateral earth pressure

H6.1Q6.1G4.1 ++

H6.1G9.0 +

H: Lateral earth pressure (Yanal toprak itkisi)

In addition to (1)

In addition to (1)



HOMEWORKİlhan Berktay, BETONARME 1, İMO İstanbul şubesi publication

Study examples : 2.4.1, 2.4.2, 2.4.3, 2.4.4 ve 2.4.5

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