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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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Analysis of Doubly Reinforced
Rectangular SectionsStrain Compatibility Check
Assume s using similartriangles
( )
( )
s
s
0.003'
' *0.003
c d c
c dc
=
=
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Analysis of Doubly Reinforced
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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Analysis of Doubly Reinforced
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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Analysis of Doubly Reinforced
Rectangular Sections
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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Analysis of Doubly Reinforced
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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Analysis of Doubly Reinforced
Rectangular Sections
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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Analysis of Doubly Reinforced
Rectangular Sections
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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Analysis of Doubly Reinforced
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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Analysis of Doubly Reinforced
Rectangular Sections
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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Doubly Reinforced Beams DesignProcedure when section dimensions
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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Doubly Reinforced Beams DesignProcedure when section dimensionsare known
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)
Doubly Reinforced Beams Design
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Doubly Reinforced Beams DesignProcedure when section dimensions
are known
6 Determine Asrequired to resist M
(assume s y)
( ) ( ) ( )s req'd y
M
A d d f
=
Doubly Reinforced Beams Design
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Doubly Reinforced Beams DesignProcedure when section dimensions
are known
7 Calculate total tension reinforcementrequired.
( ) ( )s1s req'd s req'dA A A= +
Doubly Reinforced Beams Design
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Doubly Reinforced Beams DesignProcedure when section dimensionsare known
8 Select reinforcing bars so As(provided) As(reqd)and As As(reqd). Confirm that the bars will fitwithin the cross section.
Doubly Reinforced Beams Design
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Doubly Reinforced Beams DesignProcedure when section dimensionsare known
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)
Doubly Reinforced Beams Design
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Doubly Reinforced Beams DesignProcedure when section dimensions
are known
Calculate the actual Mn for the sectiondimensions and reinforcement selected.Check strength, Mn Mu. (keep
overdesign within 10 % )
10
Doubly Reinforced Beams Design
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Doubly Reinforced Beams DesignProcedure when section dimensions
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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Example: Doubly Reinforced
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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Example: Doubly Reinforced
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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Example: Doubly Reinforced
SectionCompute the effective reinforcement ratio and
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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Example: Doubly Reinforced
SectionCompute the effective reinforcement ratio and
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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Example: Doubly Reinforced
SectionCompute the iterative values (1)
Use an iterative technique to find fs
( )( ) ( )( )( ) ( )
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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Example: Doubly Reinforced
SectionCompute the iterative values (2)
Use an iterative technique to find fs
( )( ) ( )( )( ) ( )
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
mm MPa mm MPaAsf As fsc mm
f c b MPa mm
mmf s Es MPa MPa
mm
l bl i f d
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Example: Doubly Reinforced
SectionCompute the iterative values (3)
Use an iterative technique to find fs
( )( ) ( )( )( ) ( )
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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Example: Doubly Reinforced
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
mm MPa mm MPaAsf As fsc mm
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
mm MPa mm MPaAsf As fsc mm
f c b MPa mm
mmf s Es MPa MPamm
E l D bl R i f d
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Example: Doubly Reinforced
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
= +
= +
=