L3 Analysis and Design of Rect-Section 32s-V2

Analysis and Design of Rectangular Beam Sections

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Key differences with EC2

2013 Design code EC2 (National Annex) C50

Concrete Design based on:

Cube strength Cylinder Strength

Denoted by: fcu fckDesign strength 0.87fy, 0.45fcu 0.567fckStress block depth 0.9x 0.8x

redistribution dd

kk

xbal 45.0)(

2

1

Limiting x/d 0.5 0.45Limiting z/d 0.775 0.82

Limiting K or Kbal 0.156 0.167Limiting d/d(0% redistribution)

0.19 0.171

Moment x


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Composite ActionThe strains at concrete and steel at the same level are the same !!

Effect of shrinkage and thermal movement on concrete causes cracking in concrete but strengthens the bond between concrete and steel bar

Effect of creep on concrete a change in the compressive strains transferred to reinforcing steel


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Concrete stress-strain relationship

back

Design stress = 0.67fcu/1.5 = 0.447fcu 0.45fcu

(for fcu


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Steel stress-strain relationship

back

For fy = 500 N/mm2

yy

sm

fE

m=1.15

Mild steel fy = 250 N/mm2High Yield steel fy = 500 N/mm2

Short-term design stress-strain curve for reinforcementFigure 4.2:

y k

m

f

S

t

r

e

s

s

N

/

m

m

2

Strain

Tension andcompression

200kN/mm2

y 002170y .


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Distribution of strain and stress under bending

axis

triangular rectangularparabolic

equivalentrectangular

(a) (b) (c)

Section Strains Stress Blocks

st

cc

sc

Section with strain diagrams and stress blocks

As

A' s

Figure 4.3:

neutralx

d'

d

s = 0.8xs=0.9x

effective depth

C

T

(d)

force

l

e

v

e

r

a

r

m

back


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Assumptions in the Reinforced Concrete Theory Concrete does not take any tension. (1~2 N/mm2) The stresses in concrete in compression follows the simplified

parabolic-rectangular stress-strain curve in (Concrete2013).

The ultimate limit of collapse in concrete is reached when the strain at the extreme compression fibre reaches 0.0035

The steel stresses follows the simplified short-term design stress-strain curve in (Concrete 2013).

A perfect bond exists between the steel bar and concrete. Where a section is designed for flexure only, the lever arm should

NOT be assumed to be greater than 0.95 times the effective depth.

Plane sections before bending remain plane after bending, i.e. the strain distribution of concrete in compression and steel in tension or compression are directly proportional to their distance from theneutral axis at which the strain is zero.

To

To

To

To

To

)( 2cu 60N/mmf


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2

21

and

cu

stcust

dxxxd

0.5x d

st

cu2

scAs

A' s

x

d'

d

(

d

-

x

)

2

2

for 0.0035 (concrete crushes)and 0.00217 (steel yields)

( 500 / )

cu

st y

yf N mm

To ensure yielding of the tension steel 0.6170.0021710.0035

dx d

The tension steel yields when

RC Beam Section under bending moment

Concrete2013 however places limit at ----

--- Balanced Failure ---

back

Safe ???


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st

cu2

scAs

A' s

x

d'

d

(

d

-

x

)

RC Beam Section under bending moment

Take high tensile bar for example

If x=0.617d, st= 0.002. Steel yields and concrete crushes at the same time

If x 0.002. Steel SURELY yields BEFORE concrete crushes

If x>0.617d, st


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Equivalent Bending stress block for Design

axis

Section Strains Stress Block

st

Singly reinforced section with rectangular stress block

0.0035 0.85f / = 0.567f ckck c

Fst

Fcca

s=0.8x

Figure 4.4:

As

b

d

s/2

z = l d

xneutral

0.67fcu/m=0.45fcu

s=0.9x

< 0.95ddepth of stress block

They form a couple !!


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and stress area 0.45 cuF f bscc (4.5) For equilibrium: zFzFM stcc

substitute 2/sdz (4.6) to give: 0 45 0.9 ( )cu cuM . f bs z f b d z z (4.7)

to give 2( / ) ( / ) / 0.9 0z d z d K Re-arrange the above and define

2/ cuK M bd f

We have ( / ) ( /1.15) 0.87st y m s y s y sF f A f A f A Solve for z/d to have [0.5 (0.25 /0.9)]z d K (4.8)

Substitute in (4.5) 0.87s yMAf z

(4.9)

K factor

Area of steel required

Input: M, b,d, fcu,fy

Output: As


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K=0.156 corresponds to x=0.5d and z=d-s/2=0.775d, with Mu=0.156fcubd2=Kfcubd2

K

Mu is the max. moment provided by the concrete section limited byx=0.5d or it is the moment capacity of the concrete section

compression reinforcement required to provide BM larger than Mu

Obtain this curve by plotting Eq. (4.8)

l

e

v

e

r

a

r

m

f

a

c

t

o

r

upper limit or range

lower limit or range


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The balanced sectionThe code specifies xbal = 0.5 d

Depth of the stress block, s=0.9 xbal = 0.45 d

Force in the concrete stress block,

Fccbal = 0.45 fcu bs = 0.2025 fcubdFor equilibrium of forces, Fstbal = 0.87 fy Asbal = Fccbal = 0.2025 fcubd

Asbal = 0.2328 fcubd/fywhich is the steel area in a balanced section

The ultimate moment of resistance of the balanced section,

Mbal = Fccbal zbal = 0.2025 fcubd (d - s/2) = 0.156 fcubd2

When the design moment Md = Mbal,bal2

cu

d K1560bdf

M .)( 2cu N/mm45f

)( 2cu N/mm45f


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Singly reinforced beam design: Ex 4.1 - Determine As for given M

62185 10 1200

0.87 0.87 500 369s y

MA mmf z

As

b= 260

d

=

4

4

0

kNmM

fyfcu185

2

Lever Arm:

{0.5 0.25 }0.9

0.122440{0.5 0.25 } 3690.9

Kz d

mm

2

6

2

185 10 0.122 0.156260 440 30

Compression steel not required

cu

MKbd f

Checking the sufficiency of section !

Step 1:

Step 2:

Step 3:


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Equilibrium : 0.45 0.87

0.45 25 300 0.87 500 1470 189.5 and / 0.9 189.5 / 0.9 210.5 0.5Hence steel has yielded

cc st

cu y s

F Ff bs f A

ss mm x s mm d

0.567f

ck

Fst

Fcc

s

b = 300

d

=

5

2

0

axisneutral

sA = 1470 mm

x

z

2

Singly reinforced beam design: Ex 4.2 Find the resistant moment of section given b, d and As

0.45fcu

fyfcu

To


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6

Moment of Resistance of the section:0.87 ( / 2) 0.87 500 1470(520 189.5 / 2) 10271.9

st y sM F z f A d skNm

0.567f

ck

Fst

Fcc

s

b = 300d

=

5

2

0

axisneutral

sA = 1470 mm

x

z

2

0.45fcu

The moment of resistance can also be calculated as M = Fcc z

)( 2cu N/mm45f

Note that

x = (d z)/0.45


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Rectangular section with compression reinforcement

Cross-section with compression reinforcementback

Mu

= Fst1+Fst2


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Yielding of Asc will be checked separately !

For a singly reinforced section : 20.156u cuM f bd = K fcubd2

Even though the design (applied) moment exceeds this value, we still require x0.5d to ensure steel yielding and a ductile failure Therefore: For equilibrium: scccst FFF So with the compression reinforcement at yield:

'0.87 0.45 0.87y s cu y sf A f bs f A

and with 0.9 0.5 0.45s d d

'0.87 0.2025 0.87y s cu y sf A f bd f A

/ 2 0.9 / 20.9 0.5 / 2 0.775

z d s d xd d d

The lower limit of range for z


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and taking moments about the centroid of the tension steel:

'

2 '

( ')0.2025 0.775 0.87 ( ')

0.156 0.87 ( ')

cc sc

cu y s

cu y s

M F z F d df bd d f A d d

f bd f A d d

Re-arrange to give the compression steel as:

2'

'

0.156 0.87 ( ')

cus

y s

M f bdAf A d d

2

' 0.1560.87 ( ')

cus

y

M f bdAf d d

(4.14)

(4.15)Mu


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Multiply both sides of Eq. (4.13) by z=0.775d:

and with K =0.156 and K=M/bd2fcu:

20.156 0.87

'cus s

y

f bdA Af z

2' ( ')

0.87 ( ')cu

sy

K K f bdAf d d

2'

0.87'cu

s sy

K f bdA Af z

(4.16)

(4.17)

(4.18)

To


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For 2500 /yf N mm and 0.002175sc y the

compression steel will have yielded only if:

2cuf 60N/mm' 0.0021751 0.38, when

0.0035 0.5dx x d

The Design code Concrete2013 specifies

d/x 0.38 to ensure the compression steel to yield before section fails. (The criteria can also be stated as d/d 0.19 )If d/x > 0.38, refer to Figure 3.9 in the Code for the compressive stress.

(4.20)

Check on yielding of compression steel -

,

d1


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Doubly reinforced beam design: Ex 4.4 Find moment of resistance of section

fcu=25 fy=500

M=? kNm

= Fst1+Fst2


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'

'

Equilibrium : 0.87 0.45 0.87

0.87 ( )

0.450.87 500(2410 628) 246.1

0.45 25 280with 246.1 , / 0.9 246.1/ 0.9 273.4 0.536 0.617

Hence tension steel has yi

st cc sc

y s cu y s

y s s

cu

F F Ff A f bs f A

f A As

f b

mm

s mm x s mm d d

elded

Also '/ 50 / 273.4 0.183 0.38Hence compression steel has yielded as assumed.

d x in real situation


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'

6

Taking moments of compressive forces about the tension steel( / 2) ( ')

0.45 ( / 2) 0.87 ( ')0.45 25 280 246.1(510 246.1/ 2) 0.87 500 628(510 50)425.63 10 425.63

cc sc

cu y s

M F d s F d df bs d s f A d d

Nm kNm

Section Stress Block

0.567fck

Fst

Fscs=0.8x

A = 2410s

b = 280

d

=

5

1

0

FccA' = 628

s

d' = 50

0.45fcu

s=0.9x

= Fst1+Fst2


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Doubly reinforced beam design: Ex 4.3 - Find As, As

As

b = 260

A's

d

=

4

4

0

d' = 50

25500

285

cu

y

ff

M kNm

2 2

2' 0.156 25 260 440 519.3 1842.70.87 0.87 500(0.775 440)

'cus s

y

K f bdA A mmf z

2'

22

( ')0.87 ( ')

(0.226 0.156)25 260 440 519.3

0.87 500 (440 50)

cus

y

K K f bdA

f d d

mm

K6

2 2

285 10 0.226 0.156260 440 25

Compression steel is required50 440 0 11 0 19

Compression steel has yielded

cu

MKbd f

d'/d / . .

Check the sufficiency of section !Step 1:

Step 2:

Step 3:


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Design of tension and compression reinforcement - Ex. 7.3

The beam section shown below has characteristic material strengths of fcu = 25N/mm2 and fy = 500N/mm2 The ultimate moment is 165kN m. causing hogging of the beam. Check on sufficiency of concrete section

6

2 2

165 10 0.26 ' 0.156230 330 25

cuMK K

bd f

This means compression steel is required.

Step 1:


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Design of tension and compression reinforcement

x=0.5d=165mm; d/x=50/165=0.303 < 0.38 (o.k.)

Therefore:

Compression steel:

Tension steel:

Provide two T20 bars for As=628mm2 and four T25 bars for As = 1960mm2.

0.87sc yf f

2 6 2'

2

( 0.156 ) (165 10 0.156 25 230 330 )0.87 ( ') 0.87 500 (330 50)

552.6

cus

y

M f bdAf d d

mm

2 2'

2

0.156 0.156 25 230 330 552.60.87 0.87 500 0.775 330

878.0 552.6 1430.6

cus s

y

f bdA Af z

mm

To

Step 2:

Step 3:


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Evolution of Concrete Compressive Block with Moment of Resistancein Design Rectangular Beam Section

increasing applied bending moment on section

x0.5d x


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Design Chart depends on the moment redistribution factor

over-reinforcedunder-reinforced

Margin not used in Code

Balanced point

back

For

For illustration ONLY !!


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Example 1B=300, d=450, M=262.5 kNm

Solution

2

2

6

2

51849

371100chartfrom

324450300

105262

mm.A

.bdA,

..bdM

s

s

)1960(254Provide 2mmAT s To



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Example 2B=300, d=450, M=320 kNm

Solution

)2450(255

2093

551100chartfrom

,50100ofcurvenexttheUse

27545030010320

2

2

2

6

2

mmTmmA

.bdA,

.'/bdA

.bdM

s

s

s

)943(20367550100 22 mmT,mmA,.bdA '

'

s

s To

No Good !

x/d > 0.5


~~~ End ~~~


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To

Back


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Distribution of strain and stress

axis

triangular rectangularparabolic

equivalentrectangular

(a) (b) (c)

Section Strains Stress Blocks

st

cc

sc

Section with strain diagrams and stress blocks

As

A' s

Figure 4.3:

neutralx

d'

d

s = 0.8x

Back

s=0.9x

< 0.95d

Documents

L3 Analysis and Design of Rect-Section 32s-V2