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8/14/2019 Flexural Analysis and Design of Beamns 4
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Plain & Reinforced
Concrete-1CE-313
Flexural Analysis and
Design of Beams
(Ultimate Strength Design ofBeams)
8/14/2019 Flexural Analysis and Design of Beamns 4
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a n e n orceConcrete-1
Ultimate Strength Design of Beams(Strength Design of Beams)
Strength design method is based on the philosophy of dividingF.O.S. in such a way that Bigger part is applied on loads andsmaller part is applied on material strength.
fc
0.85fc
Stress
Strain
Crushing
Strength
0.003
favg
favg = Area under curve/0.003
If fc 30 MPa
favg = 0.72 fc
1 = Average Strength/Crushing
Strength
1 = 0.72fc / 0.85 fc = 0.85
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a n e n orceConcrete-1
Ultimate Strength Design of Beams (contd)
Cc
T = Asfs
la = d a/2
N.A.
cu =0.003
StrainDiagram
ActualStress
Diagram
Internal ForceDiagram
In ultimate strength design method the section is alwaystaken as cracked.
c = Depth of N.A from the extreme compression face atultimate stage
a = Depth of equivalent rectangular stress diagram.
s
h
cd
b 0.85fc
fs
0.85fca
EquivalentStress
Diagram/Whitneys
Stress
Diagram
a/2
fs
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a n e n orceConcrete-1
Ultimate Strength Design of Beams (contd)
ActualStress
Diagram
0.85fc 0.85fc
Equivalent StressDiagram/
Whitneys StressDiagram
cCc Cca
The resultant of concretecompressive force Cc, acts at thecentriod ofparabolic stressdiagram.
Equivalent stress diagram ismade in such a way that it hasthe same area as that of actualstress diagram. Thus the Cc, willremain unchanged.
a/2
ab'0.85fcbf cav =
a'0.85fc'0.72f cc =
c'0.85f
'0.72fa
c
c =
ca 1=
l i i f d
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Plain & ReinforcedConcrete-1
Ultimate Strength Design of Beams (contd)
Factor 1
1= 0.85 for fc 30 MPa
Value of 1decreases by 0.05 for every 7 MPa
increase in strength with a minimum of 0.65
0.65'0.00714f1.064 c1 = 85.0
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a n e n orceConcrete-1
Determination of N.A. Location at UltimateCondition
CASE-I:Tension Steel is Yielding at Ultimate Condition
ys
orys ff
CASE-II:Tension Steel is Not Yielding at UltimateCondition
yf
y s
y
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a n e n orceConcrete-1
CASE-I:Tension Steel is Yielding at UltimateCondition
ysss fAfAT ==
ab'0.85fC cc =2
ada =l
abffA cys = '85.0
For longitudinal Equilibrium
T = Cc
bf
fAa
c
ys
=
'85.0 1
ac =and
Cc
T = AsfsInternal ForceDiagram
a/2
la
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a n e n orceConcrete-1
CASE-I:Tension Steel is Yielding at UltimateCondition(contd)Nominal Moment Capacity, M
ndepending on steel = T x
la
= 2Mna
dfA ys
Design Moment Capacity
=
2M bnb
adfA ys
Nominal Moment Capacity, Mn depending on concrete = Ccx la
=
2
adab0.85fc'Mn
=2
adab0.85fc'M bnb Design Moment
Capacity
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a n e n orceConcrete-1
Minimum Depth for Deflection Control
I
1
3(Depth)1
For UDL
4L
( ) 3LLDeflection Depends upon Span, end conditions, Loads and fy
of steel. For high strength steel deflection is more and moredepth is required.
Pl i & R i f d
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Plain & ReinforcedConcrete-1
Minimum Depth for Deflection Control (Contd)ACI 318, Table 9.5(a)
SteelGrade
SimplySupported
One EndContinuous
Both EndContinuous
Cantilever
300 L/20 L/23 L/26 L/10
420 L/16 L/18.5 L/21 L/8
520 L/14 L/16 L/18 L/7
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Concluded