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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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Flexural StressFlexural Stress
Example of rectangular reinforced concrete beam.
-1 $etup equilibrium.
=
=
=
==
n
css
,
/)
T0
!5.0
&T0
adM
abffA
F
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Flexural StressFlexural Stress
Example of rectangular reinforced concrete beam.
-) 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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Flexural StressFlexural Stress
Example of rectangular reinforced concrete beam.
-) ind fle,ural capacit'.
( )n
s s
/ moment arm
)
T
aA f d
= =
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Flexural StressFlexural Stress
*,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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Flexural Stress Flexural Stress
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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Flexural Stress Flexural Stress
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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Flexural Stress Flexural Stress
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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Flexural Stress Flexural Stress
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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Flexural Stress Flexural Stress
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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Flexural Stress Flexural Stress
Non-RectangularNon-Rectangular
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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Flexural Stress Flexural Stress
Non-RectangularNon-Rectangular
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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Flexural Stress Flexural Stress
Non-RectangularNon-Rectangular
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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Flexural Stress Flexural Stress
Non-RectangularNon-Rectangular
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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Flexural Stress Flexural Stress
Non-RectangularNon-Rectangular
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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ResistanceResistance
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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ResistanceResistance
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