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Reinforced and Prestressed ConcreteReinforced and Prestressed Concrete -- CC 55 --

Arh. Juergen Mayer

Arh. Santiago Calatrava

Ciudad de las Artes y las Ciencias, Valencia, Spain

Designed by Santiago Calatrava and Flix Candela

2

Rectangular Section with Compression Reinforcements

(Double Reinforced Rectangular Beams)

1) to reversal moment for the beams of a reinforced frame (the

tension reinforcement provided for the negative moment

becomes the compression steel under positive moment).

2) to increase the moment resisting capacity of the section (when is

not possible to change the dimensions of the beam).

3) to reduce the long-term deflection of an element under service

loads (when the concrete begins to creep, the compressive force

in the beam is transferred from the concrete to the steel, thus

the concrete stress is lowered and deflection due to creep is

much reduced).

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A. Design of rectangular reinforced concrete section with compression reinforcement

4

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B. Analysis of doubly reinforced

rectangular section

the resistance moment of the section:

ssccRd zFzFM += 2

x

xdcus

= 31

x

dxcus

232

=

003503 .cu =MPaGPaEs

310200200 ==

13

1 =x

d

cu

s

1

3

1 +=cu

s

x

d

13

3

scu

cu

+=

131 700

700

1020000350

00350

ss

MPa.

.

+=

+

=

the strains of the reinforcements bars are:

x

d

cu

s 2

3

2 1=

3

22 1cu

s

x

d

=

23

3

scu

cu'

=

232 700

700

1020000350

00350

ss

MPa.

.'

=

=

6

the steel in tension yields if:

the steel in compression yields if:

s

ydsE

f>1

yds fd

x

+

+==

700

700

700

700

1

s

ydsE

f>2

yds fd

x'

==

700

700

700

700

22

for tension

for compression

if the steel not yields, then the steel is in the elastic domain

and the stresses are:

== 1170011

sss E

=='

Esss

1

170022

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Example 1: The resistance moment when the tension stell yields and the compression steelis in the elastic domain:

21 scs FFF += = 0X

221 sscdyds AfxbfA +=

+=x

dAfxbfA scdyds

221 1700

2222

1 700700 dAxAfxbxfA sscdyds +=

0700700 22122 =+ dAxfAxAfxb sydsscd

0700700 22212 =

cd

s

cd

syds

fb

dAx

fb

AfAx

=

cd

sydsx

fb

AfAb

70021

1=xa

=cd

sx

fb

dAc

22 700

x

xxxx

a

cabbx

+

=2

42

8

Conclusion: the compression reinforcements is not used for calculus if :

( )2222

ddAx

dfxbM sscdRd +

=

2dx > If:

and

cds f=1

ukuds . =< 901

then

x

dxs

22 700

=

and

2dx If:

=

2

xdfxbM cdRd

then

2dx

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Example 2: double reinforced section with compression reinforcement in el astic rangeversus simple reinforced section. (C20/25, PC52)

10

cm.fb

fAx

cd

yds547

1 =

=

=

2

xdfxbM cdRd

mkN.MRd = 5422

only tension reinforcement: (Example 1 Lecture 4)

Conclusion: the increasing of the resistance moment

of a double reinforced section is small compared

with a similar section simply reinforced if the

compression reinforcement is in the elastic domain

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Example 3: singly versus double reinforced sections

singly reinforceddouble reinforced

mkN.MRd = 5422mkN.MRd = 3823

1.57cm2

0.57cm2

1.00cm2

if this difference is added to the tension reinforcement and the section is consideredas simply reinforced, then an important increasing of resistance moment is obtained:

singly reinforced

mkN.MRd = 1527

12

Example 4: if we double the reinforcing percentage of the resistance steel both in the case ofsingly and double reinforced sections, we obtain:

mkN.MRd = 5422

%.p 241=

singly

reinforced

section

mkN.MRd = 8037

%.p 472=

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mkN.MRd = 3823

Double reinforced section

mkN.MRd = 1244

the steel yields both in

tension and compression

reinforcements

mkN.MRd = 0551

the compression reinforcement

is in elastic domain (not yields)

( )222

2

ddA

xdfxbM

ss

cdRd

+

+

=

xAs1 As2( ) fyd

b fcd=

MRd b x fcd d x2

As2

fyd

d d2

( )

+

...=

14

Conclusions:

in normal reinforcing situations, the compression reinforcement not leads to a significant increase

of the resistance capacity,

to over-reinforced elements, the compression reinforcement leads to a significant increase

of the resistance capacity,

it is not economic to use compression reinforcement in a section before all the singly reinforcing

possibilities are used

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Sections with resistance reinforcements on multiple rows:

==

=c

jsjsj

t

isisicd AAfxb

11

= 0X

x

dxh i

cu

si =3

13

=x

dhE i

cu

s

si

=

= 170013

x

dh

x

dhE iicussi

x

dx j

cu

sj =3

x

dE j

cu

s

sj

=13

=

=

x

d

x

dE

jjcussj 170013

16

==

=

c

j

jsj

t

i

isicd

x

dA

x

dhAfxb

11

17001700

x - is the solution of the resulting quadratic equation

Once x calculated, the reinforcements strains and stresses values can be obtained:

for tension reinforcements:

x

dxh i

cu

si =3

=

= 1003503

x

dh.

x

dxh iicusi and

ssisi E= ifs

ydsiE

f

ydsj f= ifs

ydsjE

f

The resistance moment can be calculated with the following relation:

====

++=++=c

jsjsjsj

t

isisisiccd

c

jsjsj

t

isisiccRd zAzAzfxbzFzFzFM

1111