Rectangular Reinforced Concrete Beam

8/9/2019 Rectangular Reinforced Concrete Beam

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Lecture 5 - FlexureLecture 5 - Flexure

June 11, 2003

CVEN 444


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Lecture GoalsLecture Goals

Rectangular Beams

Safety factors

Loadng and Resstance

Balanced Beams


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Flexural StressFlexural StressThe compressive zone is modeled with a equivalent

stress block.


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Flexural StressFlexural StressThe equivalent rectangular concrete stress distribution

has what is known as a 1 coefficient is proportion ofaverage stress distribution covers.

65.0

1000

400005.0!5.0

psi4000for!5.0

c1

c1

=

=

f

f


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Flexural StressFlexural Stress

Requirements for analysis of reinforced concrete beams

"1# $tress%$train &ompatibilit' ( $tress at a point in

member must correspond to strain at a point.

")# *quilibrium ( +nternal

forces balances with

e,ternal forces


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Example of rectangular reinforced concrete beam.

-1 $etup equilibrium.

=

=

=

==

n

css

,

/)

T0

!5.0

&T0

adM

abffA

F


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-) ind fle,ural capacit'.

s s

c

s s

c

0.!5

0.!5

T A fC f ab

A fa

f b

==

=


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-) ind fle,ural capacit'.

( )n

s s

/ moment arm

)

T

aA f d

= =


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*,ample of rectangular reinforced concrete beam.

- 2eed to confirm s 3 '

( )'cs

1

s

'

'

>

=

=

=

c

cd

ac

E


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Flexural Stress Flexural Stress

Rectangular ExampleRectangular Example*,ample of rectangular reinforced concrete beam.

iven a rectangular beam

fc 4000 psi

f' 60 ksi -4 7 bars

b 1) in. d 15.5 in. h 1! in.

ind the neutral a,is.

ind the moment capacit' of the beam.


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Rectangular ExampleRectangular Examplerom equilibrium -assume the steel has 'ielded

( )( )( ) ( )

c s

)

s

c

0.!5

60 ksi ).4 in.5 in.

0.!5 0.!5 4 ksi 1) in

y

y

C T

f ba f A

f Aa

f b

==

= = =

1

.5 in.4.15) in.

0.!5

ac

= = =

The neutral a,is is


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Rectangular ExampleRectangular Example&heck to see whether or not the steel has 'ielded.

'

'

s

60 ksi0.00)07

* ):000 ksi

f = = =

&heck the strain in the steel

( )

( )

s 0.00

15.5 in. 4.15) in.0.00 0.00!) 0.000)07

4.15) in.

d c

c

=

= = >

$teel 'ielded;


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Rectangular ExampleRectangular Example&ompute moment capacit' of the beam.

( )( )

n s '

)

)

.5 in.).4 in 60 ksi 15.5 in.

)

1:7: k%in. 164.! k%ft.

a

M A f d

= =

=


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Non-RectangularNon-RectangularExampleExampleor the given beam with concreterated at fc 6 ksi and the steel is

rated at fs 60 ksi. d 1).5 in.

or a non%rectangular beam

-a 8etermine the area of the steel for a

balanced s'stem for shown area of concrete.

-b 8etermine the moment capacit' of thebeam. /n

-c 8etermine the 2


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Non-RectangularNon-Rectangular

ExampleExampleThe area of the concrete section is

or a non%rectangular beam

( ) ( ) ( ) ( )c )6 in. in. 10 in. ) in.

! in

A = +

=

The force due to concrete forces.

( )( )c c

)

0.!5

0.!5 6 ksi ! in

1:.! kips.

C f A==

=


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ExampleExample=sing equilibrium9 the area of the steel can be found

c cs s c c s

s

)

s

0.!50.!5

1:.! kips

.) in60 ksi

T C

f Af A f A Af

A

=

= =

= =


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ExampleExampleind the center of the areaof concrete area

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

i i

i

6 in. in. 1.5 in. 10 in. ) in. 4 in.

6 in. in. 10 in. ) in.

).!15! in.

y Ay

A=

+= +

=


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ExampleExampleThe moment capacit' of the beam is

( )

( )

n

1:.! kips 1).5 in. ).!15! in.

1!6: k%in. 155.75 k%ft.

M T d y= =

=


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ExampleExample&ompute the 1value

c1

4000 psi0.!5 0.05 1000 psi

6000 psi 4000 psi0.!5 0.05

1000 psi0.75

f

=

=

=


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ExampleExampleind the neutral a,is

1

5.0 in.6.67 in.

0.75

ac=

= =


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Safety ProvisionsSafety Provisions

$tructures and structural members must alwa's be

designed to carr' some reserve load above what ise,pected under normal use.


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Safety ProvisionsSafety Provisions

There are three main reasons wh' some sort of safet'

factor are necessar' in structural design.

"1# &onsequences of failure.

")# >ariabilit' in loading.

"# >ariabilit' in resistance.


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Conseuences ofConseuences of

FailureFailure

?otential loss of life.

&ost of clearing the debris and replacement of thestructure and its contents.

&ost to societ'.

T'pe of failure warning of failure9 e,istence ofalternative load paths.

< number of sub@ective factors must be considered in

determining an acceptable level of safet'.


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!aria"ility in Loa#ing!aria"ility in Loa#ing

requenc' distributionof sustained component

of live loads in offices.


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!aria"ility in!aria"ility in

ResistanceResistance>ariabilit' of the strengths of concrete and

reinforcement.

8ifferences between the as%built dimensionsand those found in structural drawings.

*ffects of simplification made in the

derivation of the members resistance!


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!aria"ility in!aria"ility in

ResistanceResistance

&omparison of

measured and

computed failure

moments based onall data for

reinforced

concrete beams

with fc3 )000 psi.


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$argin of Safety$argin of Safety

The distributions of

the resistance and

the loading are usedto get a probabilit'

of failure of the

structure.


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$argin of Safety$argin of Safety

The term

A B % $

is called the safet'margin. The probabilit'

of failure is defined asC

and the safet' inde, is

Y

Y

=

[ ]0ofobabilit'?r


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Loa#ingLoa#ing

SPECIFICATI!S

&ities in the =.$. generall' base their building

code on one of the three model codesC =niform Duilding &ode

Dasic Duilding &ode -DE&


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Loa#ingLoa#ing

These codes have been consolidated in the

)000International Building Code.

Foadings in these codes are mainl' based on

ASCE Minimum Design Loads for Buildings

and t!er Structures has been updated to


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Loa#ingLoa#ingThe loading variations are taken into

consideration b' using a series of Gload factorsH

to determine the ultimate load9 =.

( )

( ) ( ) ( )

( )

r

r

1.4

1.) 1.6 0.5 or or

1.) 1.6 0.5 1.0 or or

1.) 1.0 1.0 0.)

etc.

" D F

" D F T L # L S $

" D % L L S $

" D E L S

= +

= + + + + +

= + + += + + +="


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Loa#ingLoa#ingThe equations come from


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Loa#ingLoa#ing

The most general equation for the ultimate load9

= -/u that 'ou will see is going to beC

1.) 1.6" D L= +


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The load factors will generate the ultimate load9

which is used in the design and anal'sis of the

structural member.

/u( =ltimate /oment

/n( 2ominal /oment

( $trength Beduction actor

u nM M=


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The strength reduction factor9 9 varies from memberto member depending whether it is in tension or

compression or the t'pe of member. The code has

been setup to determine the reduction.


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'nelastic (e&avior'nelastic (e&avior

Tension Failure

The reinforcement

'ields before the

concrete crushes. Theconcrete crushes is a

secondar'

compression failure.

The beam is known as

an under#reinforced

beam.


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'nelastic (e&avior'nelastic (e&avior

Which type of failure is the most desirable?

The under#reinforced

beamis the most

desirable.fs f'

s33 '

Aou want ductilit'

s'stem deflects and

still carries load.


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(alance# Reinforcement Ratio)(alance# Reinforcement Ratio)

"al"al

bal unique value to get simultaneous c 0.00K s '

=se similar trianglesC

b

'

b cdc

00.0

=

(alance# Reinforcement(alance# Reinforcement


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(alance# Reinforcement(alance# Reinforcement

Ratio)Ratio) "al"al

( )

( ) ( )

( ) ( )

b ' b

b '

bb

' '

b s

s' '

0.00d 0.00c c

c 0.00 0.00dc0.00d 0.00

cd0.00 0.00

c *0.00 !7000

d *0.00 !7000 f

=

+ == =

+ +

= = + +

The equation can be rewritten to find cb


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Nominal $oment EuationNominal $oment Euation

The equation can be rewritten in the formC

c s '

' s

c

n s '

0.!5 ba

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Rectangular Reinforced Concrete Beam