Clase 10 y 11 - Flexión _(Vigas Doblemente Reforzadas_)

7/29/2019 Clase 10 y 11 - Flexin _(Vigas Doblemente Reforzadas_)

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Flexure


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Lecture Goals

Doubly Reinforced beams


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Analysis of Doubly Reinforced

SectionsEffect of Compression Reinforcement on the Strength

and BehaviorLess concrete is needed to

resist the T and thereby

moving the neutral axis(NA) up.

TC

fAT

=

= ys


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Analysis of Doubly

Reinforced SectionsEffect of Compression Reinforcement on the Strength

and Behavior

( )12

2sc

1

c

and

2;C

ReinforcedDoubly

2;

ReinforcedSingly

aa

a

dfAMCC

adfAMCC

ysn

ysn

bal addition of Asstrengthens.

Effective reinforcement ratio = ( )

Compressionzone

allows tension steel toyield before crushing ofconcrete.


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Reasons for Providing

Compression Reinforcement

Eases in Fabrication- Use corner bars to hold & anchor stirrups.


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Effect of Compression

ReinforcementCompare the strain distribution in two beams

with the same As


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Effect of Compression

ReinforcementSection 1: Section 2:

Addition of As strengthens compression zone so that lessconcrete is needed to resist a given value of T. NA

goes up (c2 s1).

1c

ss1

11cc1c

ss

85.0

85.085.0

bf

fAc

cbfbafCTfAT

=

====

1c

ssss2

21css

2css

1cs

ss

85.0

85.0

85.0

bf

fAfAc

cbffA

baffA

CCT

fAT

=

+=

+=

+=

=


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Doubly Reinforced Beams

Under reinforced Failure ( Case 1 ) Compression and tension steel yields

( Case 2 ) Only tension steel yields

Over reinforced Failure ( Case 3 ) Only compression steel yields

( Case 4 ) No yielding Concrete crushes

Four Possible Modes of Failure


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Rectangular SectionsStrain Compatibility Check

Assume s using similartriangles

( )

( )

s

s

0.003'

' *0.003

c d c

c dc

=

=


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Rectangular SectionsStrain Compatibility

Using equilibrium and find a

( )

( )

( )

( )

( )

s s y

c sc

s s y y

1 1 c 1 c

0.85

'

0.85 0.85

A A f

T C C a f b

A f d fa

c f b f

= + =

= = =


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Rectangular Sections

Strain Compatibility

The strain in the compressionsteel is

( )( )

s cu

1 c

y

1

0.851 0.003'

dc

f d

d f

=

=


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Strain Compatibility

Confirm

;E yss

y

ys

=f

( )( )

y y1 c

s

y s

0.851 0.003' E 200000

f ff d

d f

= =


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Rectangular SectionsStrain Compatibility

Confirm

( )

( )

( )( )

y1 c

y

1 c

y y

6000.85

' 600

0.85 600' 600

ff d

d f

f d

d f f


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If the statement is true then

else the strain in the compression steel

( ) ( )n s s y s y2

aA A f d A f d d

= +

s sf E=


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Strain

Compute the stress in the compression steel.

( )( )

1 c

s

y

0.85200000 1 0.003'f df

d f

=


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Rectangular SectionsGo back and calculate the equilibrium with fs

( )s y s sc s

c

1

s

0.85

1 600

A f A fT C C a

f b

ac

df

c

= + =

=

=

Iterate until the c value is

adjusted for the fs


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Go back and calculate the moment capacity of the beam

( ) ( )n s y s s s s2aA f A f d A f d d = +


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Limitations on Reinforcement Ratio

for Doubly Reinforced beams

Lower Limit on NSR-98 C.10.5.1

same as for single reinforce beams.

cs(min) w w

y y

1.4* *4

fA b d b df f

=


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Doubly Reinforced Beams Design

Procedure when section dimensionsare known

1Calculate controlling value for the designmoment, Mu.

2 Calculate d, since h is knownd h 60

d h 90

mm

mm

For one layer of reinforcement.

For two layers of reinforcement.

D bl R i f d B D i


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Doubly Reinforced Beams DesignProcedure when section dimensions

are known

3 Estimate a c/d value, which will cause astrain, t >0.005 and find the area As1 for a

singly reinforced section. Compute c from d

D bl R i f d B D i


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are known

4 Find nominal moment capacity provided by As1

cs1

y

0.85f baA

f

=

1f1 s1 y

2

aM A f d

=

D bl R i f d B D i


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Doubly Reinforced Beams DesignProcedure when section dimensionsare known

5 Find nominal moment capacity that must be

provided by the web.

u

f1

M

M =

D bl R i f d B D i


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If , compression steel is notrequired to resist M

u/.

If , go to step 6.

Note:

uf1

MM

=

0M

0>M

Use = 0.9 for flexure without axial

load, which will be dependent onthe strain in the tension steel.(NSR-10 Sec. 9.3)



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are known

6 Determine Asrequired to resist M

(assume s y)

( ) ( ) ( )s req'd y

M

A d d f

=



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are known

7 Calculate total tension reinforcementrequired.

( ) ( )s1s req'd s req'dA A A= +



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8 Select reinforcing bars so As(provided) As(reqd)and As As(reqd). Confirm that the bars will fitwithin the cross section.



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9 Confirm that s y. If not go back to step6 and substitute fs = Es s for fy to get correct

value of As(reqd)



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are known

Calculate the actual Mn for the sectiondimensions and reinforcement selected.Check strength, Mn Mu. (keep

overdesign within 10 % )

10



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are known

Check whether provided is withinallowable limits.11

E l D bl R i f d


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Example: Doubly Reinforced

SectionGiven:

fc= 28 MPa fy = 420 MPa

As = 2 #6 As = 4 #7

d= 60 mm d = 390 mm

h=450 mm, b = 300 mm

Calculate Mn for the section for the givencompression steel.


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Section

Compute the reinforcement coefficients, the areaof the bars #7 (387 mm2) and #6 (284 mm2)

( )

( )

( ) ( )

( ) ( )

2 2

s

2 2

s

2

s

2

s

4 387 mm 1548

2 248 mm 568 mm

1548 mm0.0132

300 mm 390 mm

568 mm0.0049

300 mm 390 mm

A mm

A

A

bd

A

bd

= =

= =

= = =

= = =


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SectionCompute the effective reinforcement ratio and

minimum

( )

y

c

y

min

0.0132 0.0049 0.0083

1.4 1.4 0.0033420

28

or 0.00314 4 420

0.0132 0.0033 OK!

eff

f

f

f

= = =

= = =

= =


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minimum

Compression steel has not yielded.

( )( ) ( )

( ) ( )

1 0,85 ' ' 0,85 0,85 28 60600 600'

600 420 390 420 600 4200, 0083 0, 0247

y

f c d

d fnot true


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minimum

Use an iterative technique to find fs

( )( )

( ) ( )

1

1

0,85 ' '600 ' 600

( ')

0,85 ' ' 0,85 0,85 28 60' 600 1 600 1 64.32( ') 0,0083 390 420

y

y

f c dfs

df

f c dfs MPadf

=

= = =


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SectionCompute the iterative values (1)


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

2 2

1

1548 420 568 64.32' '101

0,85 ' 0,85 28 0,85 300

60

' ' (200000 ) 1 0,003 244101

y

s

mm MPa mm MPaAsf As fsc mm

f c b MPa mm

mm

fs Es MPa MPamm

= = =

= = =


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

2 2

1

1548 420 568 244' '84

0,85 ' 0,85 28 0,85 300

60' ' (200000 ) 1 0,003 171

84

= = =

= = =

y

s


f c b MPa mm

mmf s Es MPa MPa

mm

l bl i f d


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

2 2

1

1548 420 568 171' ' 910,85 ' 0,85 28 0,85 300

60' ' (200000 ) 1 0,003 204

91

= = =

= = =

y

s

mm MPa mm MPaAsf As fsc mmf c b MPa mm

mmf s Es MPa MPa

mm

E l D bl R i f d


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SectionCompute the iterative values (etc)

Use fs=196 MPa and c=89 mm

( )( )

( )( )

( ) ( )

2 2

1

1548 420 568 204' '88

0,85 ' 0,85 28 0,85 300

60' ' (200000 ) 1 0,003 191

88

= = =

= = =

y

s


f c b MPa mm

mmf s Es MPa MPa

mm

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

2 2

1

1548 420 568 191' '89

0,85 ' 0,85 28 0,85 300

60' ' (200000 ) 1 0,003 19689

= = =

= = =

y

s


f c b MPa mm

mmf s Es MPa MPamm

E l D bl R i f d


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SectionCompute the moment capacity of the beam

( ) ( )

( )( ) ( )( )( )

( )( )( )

3 2 3 2

3 2

' ' ' ' '2

0,85 0,0891,548 10 420000 0,568 10 196000 0,39

2

0,568 10 196000 0,39 0,06

226.5

y

aMn Asf As f s d As f s d d

mMn m KPa m KPa

m KPa m m

Mn KN m

= +

= +

=

Clase 10 y 11 - Flexión _(Vigas Doblemente Reforzadas_)

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Transcript of Clase 10 y 11 - Flexión _(Vigas Doblemente Reforzadas_)