REINFORCED CONCRETE 1 - yarbis.yildiz.edu.tr · Betonarme 1 Ders Notları Yrd.Doç.Dr.Murat Serdar Kırçıl Cross-sectional design based on the limit state design assumptions is

REINFORCED CONCRETE 10423522

Dr. Murat Serdar Kırçıl

İNM 1-024

www.yildiz.edu.tr/~kircil

ULTIMATE STRENGTH OF MEMBERS SUBJECTED TO PURE BENDING(Basit eğilme etkisindeki betonarme elemanların taşıma gücü)

The RC beams and RC slabs are subjected to flexure because of theloads applied to structures. They are also subjected to shear force in

addition to flexure. However, those actions are considered separately for

design of the RC structures since it is assumed that they are independent

from eachother.

Reinforced Concrete 1 Lecture Notes

Dr.Murat Serdar Kırçıl

This part of section is

elongating since it is

loaded by tensile

force.(Tension side)


under the effect of

compressive force and it

is shortening. (Compression side)

+

Reinforced Concrete 1 Lecture Notes

Dr.Murat Serdar Kırçıl

Betonarme 1 Ders Notları

Yrd.Doç.Dr.Murat Serdar Kırçıl

Where should the reinforcing bars

be placed?

This part of the section is loaded by tensile load. The tensile reinforcement is required for this area to compensate the tensile strength deficiency of concrete.


under the effect of

compressive force and it

is shortening. (Compression side)


elongating since it is

loaded by tensile

force.(Tension side)



+

-

Beam zones along which

reinforcement steel is

required (theoretical)

Provided reinforcement

steel



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.

Limit state design (Limit durumlara göre hesap)

Fundamental principles and methods for design of reinforced concrete structures

Main assumptions of this method



M

+

M

FC FC

FT FT



M

+

M

FC FC

FT FT

The position of the tensile force doesn’t change. However, the level of the compressive force can change.

Some books use C and T to

denote compressive force and

tensile force, respectively.

FC: Compressive force (basınç kuvveti)

FT: Tensile force (çekme kuvveti)



FC

FT

How does the

compressive stress

distribute?

The position and the value of the compressive force completely depends

on the distribution of the compressive stress since it is equal to

summation of compressive stress.

ASSUMPTION: The stress block of the concrete in compression is assumed to be same as the σ-Ɛ curve obtained experimentally from uniaxially loaded specimen.

σσσσ

εεεε

0.85fcd

0.002 0.003

Streess-strain curve given by TS500/2000.


Yrd.Doç.Dr.Murat Serdar KırçılFlexural behavior of underreinforced RC beams (Az-normal donatılı kirişler)

ab

a bσ

ε

σ

ε

a) All the tensile force is taken by the reinforcement steel as concrete in the tension

cracks.

b) If beam is underreinforced then the tensile force of reinforcement increases with

increasing bending moment and reinforcement reaches its yielding deformation.

(ε≥≥≥≥εsy,σs=fy). Reinforcement keeps elongating beyond the yielding point without any

increase in its tensile force (assumption of elastoplastic behavior – it takes moretensile force in real because of the strain-hardening). Neutral axis moves up

since the section remains plane and the depth of compression zone decreases. It

causes a change in the shape of compressive stress block. The distribution of

compressive stress is not linear anymore.

σ

εεsy

εεsy

σ



abcd

a b c d

σ

ε

σ

ε

σ

ε

σ

ε0.003σ

εεsy

εεsy

σ

εεsy

σ

εεsy

σ

0.85fcd 0.85fcd

c) The compressive stress increases rapidly with decreasing depth of neutral axis

and it reaches 85% of cylinder compressive strength. However, failure is not

observed since the adjacent fibers are not fully stressed (redistribution). At this

stage, concrete in the tension side of section cracks since it can NOT elongate as

much as the reinforcement steel.

d) The level of maximum stress moves to adjacent fibers (redistribution). Finally

failure is occured when the extreme fiber’s deformation reaches ultimate strain

(εcu=0.003).

Flexural behavior of underreinforced RC beams (Az-normal donatılı kirişler)



abcd

a b c d

σ

ε

σ

ε

σ

ε

σ

ε0.003σ

εεsy

εεsy

σ

εεsy

σ

εεsy

σ

0.85fcd 0.85fcd

OBSERVATION:The failure of RC sections are occurred when the ultimate strain of concrete is

reached (εcu=0.003). In the underreinforced beams, the reinforcement steel yieldsbefore extreme concrete fiber in compression reaches its ultimate strain capacity(the yielding of reinforcement steel is observed before concrete is crushed). The

failure is NOT sudden since reinforcement yields prior to crushing of concrete. The

large deformations and cracks are observed. This type of failure is called tensionfailure (çekme kırılması) or ductile failure.

Flexural behavior of underreinforced RC beams (Az-normal donatılı kirişler)


Yrd.Doç.Dr.Murat Serdar KırçılFlexural behavior of overreinforced RC beams (Çok fazla donatılı kirişler)

a b

a bσ

ε

σ

ε

a) All the tensile force is taken by the reinforcement steel as concrete in the tension

cracks.

b) The reinforcement steel stress is low since the area of reinforcement is

high. The depth of neutral axis will increase so that compatibility condition

can be satisfied.

σ

εεsy

εεsy



a b c d

a b c dσ

ε

σ

ε

σ

ε

σ

ε0.003σ

εεsy

c) The compressive force increases with the increasing area of compression stress

block. It also causes an increase in neutral axis depth. The reinforcement steel has

not yielded yet.

d) Finally failure is occured when the extreme fiber’s deformation in compression

reaches ultimate strain (εcu=0.003). The failure is sudden and brittle. This type of

failure is called Compression Failure (Basınç kırılması) or Brittle Failure (Gevrekkırılma) since it’s not possible to see visible cracks before failure.

εεsy

εεsy

εεsy

Flexural behavior of overreinforced RC beams (Çok fazla donatılı kirişler)



a b c d

a b c dσ

ε

σ

ε

σ

ε

σ

ε0.003σ

εεsy

OBSERVATION: The failure of RC sections are occurred when the ultimate strain

of concrete is reached (εcu=0.003). In the overreinforced beams, the extremeconcrete fiber in compression reaches its ultimate strain capacity prior toyielding of reinforcement. The failure is sudden and brittle. Design codes obligates

some limitations on section dimensions and reinforcement amount so that the brittle

failure of RC elements can be prevented.

εεsy

εεsy

εεsy

Flexural behavior of overreinforced RC beams (Çok fazla donatılı kirişler)



How can design

engineer control the

failure type of RC beam

sections?

The failure type of a reinforced concrete beam section can be controlled by limiting the reinforcement steel area.



x

N.Ad

εs

εcFC

h

bw

∫=x

0

wC dx)x(bF σσσσεc

x

εs

FC

εc= Strain of concrete (shortening)

εs=Strain of reinforcement steel

(elongation)

εc



x

N.A

d

εs

εc

σc

εc1

εc1 σc1

σc1

σc2

0.85fcd

σc3

εc2

εc2 σc2

εc3

εc3εcu

0.85fcd

εcu

σc3

T.E



x

N.Ad

εs

εc

FC

h

bw

∫=x

0

wC dx)x(bF σσσσ

Is there simpler

way to calculate

the compression

force instead of

integration?

EQUIVALENT RECTANGULAR STRESS

BLOCK

(Eşdeğer dikdörtgen gerilme bloğu)



0.85fcd

a

a

x

0.85fcd

x

εs

εc

0.85fcdba



x

k3fc

Neutral

axis

k2x

b bxfkkbxF c31mc == σσσσ

bxF mc σσσσ=

max1m k σσσσσσσσ =

c3max fk=σσσσ

σmax

maxm1 /k σσ=

cmax3 f/k σ=



x

k3fc

Neutral

axis

k2x

b

σmax

k1= The ratio between mean and maximumstress

k2= The ratio between resultant force depth and neutral axis depth

k3=The ratio between max. compressive

stress and strength of concrete



a

0.85fcd

x

0.85fcd

Yapılan deneyler, kolonların; çeliğin akma gerilmesine, betonun da silindir basınç

dayanımının yaklaşık 0.85’ine ulaştığında kırıldığını göstermektedir.

k3=0.85 k1k3=0.72

k1k3=0.72-(fck-25)0.0051

fck<25MPa

fck>25MPa

k1=0.72/0.85=0.85

k1=0.85-(fck-25)0.006

k1k3fcdbx

k2x a/2

k2x=a/2

0.85fcdba

b b

k1 k3 fcd b x = 0.85 fcd b a

k1 0.85 fcd b x = 0.85 fcd b a k1 x = a

k2x=a/2k2 =k1/2

k2=0.425

k2=0.425-(fck-25)0.003

fck<25MPa

fck>25MPa



ax

a/2a/2

∫=x

0

wC dx)x(bF σσσσ

0.85fcd

σσσσ

εεεε

0.85fcd

0.002 0.003

0.85fcd

abf85.0F wcdC =

0.85fcd

a

bw



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

reinforcement.

Bernouilli-Navier hypothesis is valid. Plane sections remain plane after

bending.

ASSUMPTIONS

The stress-strain relationship for

reinforcement steel is elastoplastic. fy

Concrete σ-ε curve is the curve obtained from uniaxial compression.

x

σ

ε

0.85fcd

0.002 0.003




Rectangular sections

Sections with rectangular compression zone are called rectangular section

M M

M

Rectangular beams reinforced for

tension only / single reinforced

rectangular beams (Tek donatılı

dikdörtgen kesitler)

M

Double reinforced rectangular

beams (Çift donatılı





M

Rectangular beams reinforced

for tension only (Tek donatılı


M

Double reinforced rectangular

beams (Çift donatılı dikdörtgen

kesitler)

Tensile force is carried bytensile reinforcement /Compressive force iscarried by concrete incompression zone

Tensile force is carried by tensilereinforcement / Compressive forceis carried by concrete incompression zone andcompressive reinforcement

Compression reinforcement

Tension reinforcement

M


Compression reinforcement

Rectangular sections

ULTIMATE STRENGTH OF MEMBERS SUBJECTED TO PURE BENDING




GEOMETRİ

b

bw

d

d’

h

As’

As

c

c

As: Tensile reinforcement (Çekme donatısı)

As’: Compressive

reinforcement (Basınç

donatısı)

bw: Beam width (Kiriş genişliği)

h: Beam height (Kiriş yüksekliği)

b: Effective beam width (Etkili tabla genişliği)

c: Cover concrete (Kabuk

betonu)

d’: ? ( Paspayı)

d: Effective beam depth (Etkili derinlik / faydalı derinlik )



T.Ed

εs

εc

h

bw

0FF TC =−

xFC

FT

a

0.85fcd

a/2FC

FT

d-0.5aM

The resultant of internal forces is 0 since the section is subjected to bending

moment only.

sswcd Aabf85.0 σ=

TC FF =

Force equilibrium equation

As

σs= Tensile stress in reinforcement sss E ε=σ

Es= Modulus of elasticity of

reinforcement

25

s mm/N10.2E =

fy

εεsy

σs

Es



T.Ed

εs

εc

h

bw

0FF TC =−

xFC

FT

a

0.85fcd

a/2FC

FT

d-0.5aM

sswcd Aabf85.0 σ=

TC FF =

As

+ The tensile force carried by concrete

ASSUMPTION: Concrete doesn’t carry tensile force. All tensile force carried by

reinforcement steel.


moment only.



0FF TC =−

sswcd Aabf85.0 σ=

TC FF =

εs

εc

T.Ed

h

bw

xFC

FT

a

0.85fcd

a/2FC

FT

d-0.5aM

As

Moment equilibrium equation

Mr= Kesitin taşıma gücü momenti (moment taşıma gücü)

)a5.0d(abf85.0M wcdr −=)a5.0d(A ss −σ=

)a5.0d(F)a5.0d(FM TCr −=−=


moment only.

Force equilibrium equation



Failure types

εs

εc=0.003

T.Ed

h

bw

xFC

FT

a

0.85fcd

a/2FC

FT

d-0.5aM

As

In this section, it is assumed that beams fail with bending moment effect since their shear capacity is higher than their moment capacity.

The failure criteria is the strain level at the outermost fiber: When the outermost

fiber strain reaches concrete crushing strain (εεεεcu= 0.003) RC section fails.

The parameter which controls the failure type is reinforcement steelpercentage.



Failure Types

εs

εc

T.Ed

h

bw

xFC

FT

a

0.85fcd

a/2FC

FT

d-0.5aM

As

Tension failure (çekme kırılması / sünek kırılma, düktil kırılma)

In underreinforced beams the tensile reinforcement reaches its yield deformation prior tocrushing of concrete at the outermost fiber in compression. The large deformations andcracks are observed during the failure.

Balanced failure (Dengeli kırılma)

Reinforcement steel yields and concrete is crushed simultaneously. The failure is sudden andbrittle. This type of failure is borderline case between tension and compression failure.

Compression failure (basınç kırılması / gevrek kırılma )

In overreinforced beams the concrete in outermost fiber reaches crushing strain capacityprior to yielding of tensile reinforcement. The failure is sudden and brittle.



Failure Types

εs

εc

T.Ed

h

bw

xFC

FT

a

0.85fcd

a/2FC

FT

d-0.5aM

As

Tension failure

Balanced failure

Compression failure

bρ<ρ

bρ=ρ

bρ>ρ

ρ: Reinforcement ratio

ρb: Balanced

reinforcement ratio

The parameter which controls the failure type is reinforcement steel ratio.



sswcd Aabf85.0 σ=TC FF =

εs

εc

T.Ed

h

bw

xFC

FT

a

0.85fcd

a/2FC

FT

d-0.5aM

As

)a5.0d(abf85.0M wcdr −=)a5.0d(fA yds −=

Tension failureReinforcement yields prior to crushing of concrete

0.85fcd

0.002 0.003

sys ε≥ε

fyd


?s =σ

yds f=σydswcd fAabf85.0 =

wcd

yds

bf85.0

fAa =

Material and section properties are known, ultimate moment capacity is calculated.



sswcd Aabf85.0 σ=

TC FF =

εs

εc

T.Ed

h

bw

xFC

FT

a

0.85fcd

a/2FC

FT

d-0.5aM

As

Compression failure(1) Concrete is crushed prior to yielding of reinforcement

0.85fcd

0.002 0.003

sys ε<ε

fyd


?s =σ

yds f<σ

? ?

εs

0.003

xd

x

xd

003.0

s −=

ε

x

xd003.0s

−=ε

Compatibility equation


ASSUMPTION: Plane sections

remain plane after bending.



sswcd Aabf85.0 σ=TC FF =

εs

εc

T.Ed

h

bw

xFC

FT

a

0.85fcd

a/2FC

FT

d-0.5aM

As

0.85fcd

0.002 0.003

fyd


? ?

εs

0.003

xd

x

xd003.0s

−=ε

sss E ε=σ

MPa10.2E 5

s =

x

xd600s

−=σ

85.0k1 =

1k

ax =

)1a

dk(600 1

s −=σ

dhd ′−=

xka 1=

Concrete is crushed prior to yielding of reinforcement


Compression failure(2)



)1a

dk(600 1

s −=σ

)1a

dk(600Aabf85.0 1

swcd −=

TC FF =

εs

εc

T.Ed

h

bw

xFC

FT

a

0.85fcd

a/2FC

FT

d-0.5aM

As

)a5.0d(abf85.0M wcdr −=

)a5.0d(A ss −σ=

0.85fcd

0.002 0.003

fyd


0dkA600a)A600(a)bf85.0( 1ss

2

wcd =−+Solve the second

order equation to

find a

s

ss

E

σ=ε

Compression failure(3)



ydbsbwcd fAabf85.0 =

TC FF =

εs

εc

T.Ed

h

bw

xFC

FT

a

0.85fcd

a/2FC

FT

d-0.5aM

As

Balanced failure(1) Reinforcement steel yields and concrete is crushed simultaneously (Material and section properties are known, ultimate moment capacity is calculated)

0.85fcd

0.002 0.003

sys ε=ε

fyd


?s =σ

d

x

003.0

003.0 b

ys

=ε+

yds f=σ

εsy

0.003

xb

d

sy

b003.0

003.0dx

ε+=

b1b xka =

yd

1bf600

600dka

+=

wcd

ydbs

bbf85.0

fAa =

s

yd

syE

f=ε



εs

εc

T.Ed

h

bw

xFC

FT

a

0.85fcd

a/2FC

FT

d-0.5aM

As

0.85fcd

0.002 0.003

fyd


yd

1

wcd

ydbs

f600

600dk

bf85.0

fA

+=

ydyd

wcd1sb

f600

600

f

bf85.0dkA

+=

ydyd

cd1

w

sb

f600

600

f

fk85.0

db

A

+=

db

A

w

sbb =ρ

ydyd

cd1b

f600

600

f

fk85.0

+=ρ

Balanced reinforcement ratio. It’s

function of material properties only




ydbsbwcd fAabf85.0 =

εs

εc

T.Ed

h

bw

xFC

FT

a

0.85fcd

a/2FC

FT

d-0.5aM

As

0.85fcd

0.002 0.003

fyd


wcd

ydbs

bbf85.0

fAa =

dbA wbsb ρ=

)a5.0d(abf85.0M bbwcdrb −=)a5.0d(fA bydsb −=




0.003

N.Ad

h

bw

Tension failure

xb

εs=εsy

εs<εsy

εs>εsy

x<xb x>xb

Balanced failure

Compression failure



mm40dddh +=′+=d

~20-25mm~8-10mm

~8-10mm



Design of single reinforced concrete sections (kesit hesabı)

Md Md

Parameters to be calculated: As (or ρ)

Design of RC beam sectionsThe probability of reaching ultimate limit state (failure) must be limited with a prescribed level

Section must exhibit ductile behavior even it reaches ultimate limit state (failure)

The probability of reaching serviceability limit state must be prevented (excessive deflection, crack, vibration etc.)

Loads are increased by load coefficients while material strength is decreased by

material coefficients (partial safety factors)

We use upper and lower limits for reinforcement ratio

We use upper and lower limits for reinforcement ratio. Also section height can be

limited.

Known parameters: Md, fcd, f yd, bw, h



Md Md

UPPER LIMITS

bm 85.0 ρ=ρ≤ρDuctile behavior limit (ductility requirement): (TS500)

Deflection limit:

yd

cd

f

f235.0≤ρ (Recommendation)

Earthquake zone limit (TS500)

(redistribution requirement)bd 4.0 ρ=ρ≤ρ

Max. redistribution percentage %15bd 6.0 ρ=ρ≤ρ

The max reinforcement ratio can never be more than 0.02

02.0≤ρTS500 and Earthquake

Specifications



Max. redistribution percentage %10




Md Md

Limits for reinforcement steel (Lower limit)

Concrete in tension zone will carry tensile load until it cracks. Assoon as it cracks, the all tensile force will be transferred to thereinforcement steel. Amount of the provided reinforcement steelmust be enough; so that, aforementioned transferred load can becarried. Otherwise we can not prevent brittle failure.

Minimum reinforcement ratio is obtained by equating the moment capacity of both cracked and uncracked beam section.






Assumptions

yI

M=σ

y

IM

σ=

y

IfM ctf

cr = ctkctf f2f ≈

fctf= Tensile strength in flexure

fctk= Tensile strength in direct

tensiony

If2M ctk

cr =The cracking momentof a beam withoutreinforcement

The moment of inertia of a T- beam can be assumed as approximately 1.5 times that of a rectangular beam.

12

bh5.1I

3

≈

yThe length between neutral axis and outermost tension fiber is assumed to be 0.6h.

2

ctk

3

ctkcr bhf417.0

12h6.0

bh5.1f2M ==

h6.0y ≈






Assumptions

)a5.0d(fAM yks −=The ultimate strength of a

reinforced concrete beam

The effective depth is assumed to be 0.9h

h9.0d ≈

Moment arm is assumed to be 0.85d

0.9h

0.85dd85.0a5.0d ≈−

hfA765.0)h9.0(85.0fAM yksyks ==






2

ctk

3

ctkcr bhf417.0

12h6.0

bh5.1f2M ==

hfA765.0)h9.0(85.0fAM yksyks ==

h9.0d ≈

hfA765.0bhf417.0 yks

2

ctk =

yk

ctk

yk

ctks

f

f545.0

f

f

765.0

417.0

bh

A==

5.1

ff ctk

ctd =15.1

ff

yk

yd =

yd

ctds

f

f71.0

bh

A=

bd

As=ρ

bh

A9.0 s=ρ

yd

ctd

f

f789.0=ρ

yd

ctdmin

f

f8.0====ρρρρ

Minimum reinforcement ratio TS500


Limits for reinforcement steel (Upper limit)




Tensile reinforcement is assumed to be yielded!

Md

bw

h d

ydswcd fAabf85.0 =

yds f=σ

)a5.0d(fAM ydsd −=

wcd

yds

bf85.0

fAa =

)bf85.0

fA5.0d(fAM

wcd

yds

ydsd −=

0M)bf(A)dbff(A)f59.0( dwcdswcdyd

2

s

2

yd =+− sA






Md

bw

h d

0M)bf(A)dbff(A)f59.0( dwcdswcdyd

2

s

2

yd =+−

The validity of yielding assumption is checked

db

A

w

s=ρ

???maxρρρρρρρρ ≤

Check upper and lower limits of reinforcement by using

requirements given by code. If they are satisfied then thecalculated reinforcement is OK. Select diameter and

number of reinforcement bars.

The failure type is compression and the beam section can not exhibit ductile behavior. The section height can

be increased or compressive reinforcement can be used

(double reinforced beam section).

maxρ≤ρ

maxρ>ρ

sA






Md

bw

h d

Upper limits for reinforcement

bm 85.0 ρ=ρ≤ρDuctile behavior limit

Sehim denetimi sınırıyd

cd

f

f235.0≤ρ

Redistribution (requirement

for earthquake regions)bd 4.0 ρ=ρ≤ρ Maximum redistribution %10

Maximum redistribution % 15bd 6.0 ρ=ρ≤ρ






The minimum diameter of reinforcement bar is 12mm.

The minimum number of reinforcement bar is 2 (so

that transverse reinforcement can be connected with

longitudinal reinforcement)

Tables, which shows the required minimum beamwidth so that all bars can be placed at the samelevel, are available in some textbooks. [Betonarme 1

(İlhan Berktay)]


The selected reinforcement must be checked if itcan be placed at the same level or not. If not, some

reinforcement bars can be placed at a second level.

The center of gravity and the effective depth arechanged.

Documents

REINFORCED CONCRETE 1 - yarbis.yildiz.edu.tr · Betonarme 1 Ders Notları Yrd.Doç.Dr.Murat Serdar Kırçıl Cross-sectional design based on the limit state design assumptions is