Flexural Analysis and Design of Beamns 3

8/14/2019 Flexural Analysis and Design of Beamns 3

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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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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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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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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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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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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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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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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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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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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

Documents

Flexural Analysis and Design of Beamns 3