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Plain & Reinforced
Concrete-1CE-313
Flexural Analysisand Design of
Beams
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a n e n orceConcrete-1
Minimum Reinforcement of Flexural Members(ACI 10.5.1)
As min = fc / (4fy) x bwd 1.4/fy x bwdCritical if fc < 28 Mpa
The minimum steel is always provided in structural membersbecause when concrete is cracked then all load comes on steel, sothere should be a minimum amount of steel to resist that load toavoid sudden failure.
For a design to be safe M < Mr
For an economical design M = Mr
M = Asfs x jd
As= M / (f
sx jd)
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a n e n orceConcrete-1
Minimum Depth of SectionTo satisfy a cross section against concrete failure
M = Mr
M = Ccx jd
M = (fc/2 x kd x b) x jd
dmin = (2M / fc k j b)
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Plain & ReinforcedConcrete-1
Determination of k value
k = (n2
2
+ 2n) n
Value of k can not be determined as is not know. There aretwo different approaches to establish the value of k.
1. Simultaneous Occurrence of Maximum Permissible Steeland Concrete Stresses
2. Assuming some suitable steel ratio
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a n e n orceConcrete-1
Determination of k value (contd)1- Simultaneous Occurrence of MaximumPermissible Steel and Concrete Stresses
c
Strain Diagram
s
kd
d- kdd
B C
A
E D
N.A.
Consider ABC & ADE
s/(d-kd) = c/(kd)
(fs / Es) / (d-kd) = (fc/Ec) / (kd)
fs
x kd = (Es
/Ec
) x fc
x (d-kd)
fs x k = n x fc x (1-k)
fs x k + nfck = nfc
k = nfc / (fs + nfc)
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a n e n orceConcrete-1
Determination of k value (contd)
2- Assuming some suitable steel ratioIn this approach, some suitable value of steel ratio (less than max ) is selected at the start of calculations and used for the
determination of k
max =0.75 [0.85 x 1 x (3/8) x (fc/fy)]
max = 0.75[0.85 x 1 x (fc/fy) x (600/(600
+ fy)]
Calculate max and select some value less than this.
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a n e n orceConcrete-1
Reduced Moment of Inertia Due to CrackingElastic Deflection can be expressed in general form
= f (loads, spans, supports) / ( EI )
Deflection of a cracked section can be expressed as
= f (loads, spans, supports) /( EIe)
As the number of cracks and their width increases the M.O.I ofcross section reduces and deflection increases.
Ie
= Effective moment of inertia of cracked section
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a n e n orceConcrete-1
Reduced Moment of Inertia Due to Cracking(contd)
Ie = [ Mcr /Ma ]3 Ig + [ 1 {Mcr /Ma}
3 ]Icr < IgWhere
Mcr = Cracking moment Mcr = fr Ig/yt
Ma = Maximum moment in the member at a load level where we want
to calculate deflection
Ig = Gross moment of inertia of un-cracked transformed section
Icr = Moment of inertia of cracked transformed section
fr = Modulus of rupture
yt = Extreme tension fiber distance from N.A.
Effective moment of inertia calculated by the above expression can beused to calculate the deflection of beams after cracking
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a n e n orceConcrete-1
Long Term DeflectionLong term deflection is caused by creep and shrinkage
t = i
wheret =Long term deflection
i = Instantaneous Deflection
= Multiplier for additional deflection due to long term effect
Total Deflection = i +t
Total Deflection = i + i
Total Deflection = (1
+
) i
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a n e n orceConcrete-1
Long Term Deflection (contd)
= / (1 + 50 ) = compression steel ratio
= Area of compression steel / (b x d)
= As / (b x d)
= Time dependent factor for sustained loads
Elapsed Time
5 years or more 2.0
12 months 1.4
6 months 1.2
3 months 1.0
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a n e n orceConcrete-1
Example (un-cracked transformed section)A rectangular beam of size 250mm x 650mm with effectivedepth of 590mm and is reinforced with 3 # 25 US customarybars. Concrete has specified compressive strength 28MPa.Yield strength of steel is 420 MPa. Modulus of rupture (tensile
strength) is 3.5 MPa. Determine the stresses caused by thebending moment of 50kN-m.
Datafc = 28 MPafy = 420 MPa
fr = 3.5 MPaM = 50 kN-m(positive moment)
ftop = ? fbottom = ?
b =250mm
d =590mm
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(n-1)As/2(n-1)As/2
TransformedSection
'0.5r cf f= If not given directly
2
'
3 510 1530
200,000 , 4700 24870
8
s
s c c
s
c
A mm
E MPa E f MPa
En
E
= =
= =
= =
Assume: stresses are within elastic range;No cracking at the given moment level
( )
( )( ) ( ) ( )
2
3 22 4 4
1 10710
650250 650 10710 5902 341
250 650 10710
250 650 650250 650 10710 590 642700 10
12 2
sn A mm
y mm
I y y mm
=
+ =
+
= + +
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a n e n orceConcrete-1
Example (Cracked transformed section)A rectangular beam of size 250mm x 650mm with effectivedepth of 590mm and is reinforced with 3 # 25 US customarybars. Concrete has specified compressive strength 28MPa.Yield strength of steel is 420 MPa. Modulus of rupture (tensile
strength) is 3.5 MPa. Determine the stresses caused by thebending moment of 120kN-m.
Datafc = 28 MPafy = 420 MPa
fr = 3.25 MPaM = 120 kN-m(positive moment)
ftop = ? fbottom = ?
b =250mm
d =590mm
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( ) ( ) ( )
'
2 2
4700 24870
As 15308, 0.01037
b x d 250 590
0.08298
2 0.08298 2 0.08298 0.08298
0.333
1 0.8893
c c
s
c
E f MPa
En
E
n
k n n n
kj
= =
= = = = =
== + =
=
=
b
CrackedTransformed
Section
kd
nAsd
Suppose section is cracked and material is withinelastic range
( )
( )
3 22
.
4 4
6
250 196.5 196.5
250 196.5 12240 590 196.512 2
252,750 10
120 10, 149.5
1530 0.889 590
N A
s s s
s
I
mm
MM A f jd f MPa
A jd
= + + =
= = = =
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Steel is within elastic range
( )
6
22
2 2 120 109.32
0.333 0.889 250 590c
M f M
kjbd
= = =
( ),0.45 12.6c cpermissible f f MPa= =
( )6
( ) 4
120 10 650 196.521.5
252750 10ct bottom r
My f MPa
I
= = = >>
( )0.4 168s ypermissible f f MPa= =
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Example
Design a rectangular section for a simplysupported beam of 5 m span subjected touniformly distributed load of 50kN/m, using
ASD. fc=30MPa, and fy=420 MPafc(permissible)=0.45fc,
fy(permissible)=0.4fy
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Concluded