1

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).

A. Design of rectangular reinforced concrete section with compression reinforcement

3

4

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.

.'

5

• the steel in tension yields if:

• the steel in compression yields if:

s

yds E

f1

yds fd

x

700

700

700

700

1

s

yds E

f2

yds fd

x'

700

700

700

700

22

for tension

for compression

• if the steel not yields, then the steel is in the elastic domainand the stresses are:

11

70011 sss E

'

Esss 1

170022

6

Example 1: The resistance moment when the tension stell yields and the compression steel is 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

1xa

cd

sx fb

dAc

22 700

x

xxxx

a

cabbx

2

42

7

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

2222ddA

xdfxbM sscdRd

2dx • If:

and

cds f1

ukuds . 901

then

x

dxs

22 700

and

2dx • If:

2

xdfxbM cdRd

then

2dx

8

Example 2: double reinforced section with compression reinforcement in elastic range

versus simple reinforced section. (C20/25, PC52)

9

cm.fb

fAx

cd

yds 5471

2

xdfxbM cdRd

mkN.M Rd 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

10

Example 3: singly versus double reinforced sections

singly reinforceddouble reinforced

mkN.M Rd 5422mkN.M Rd 3823

1.57cm2 –0.57cm2

1.00cm2

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

singly reinforced

mkN.M Rd 1527

11

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

mkN.M Rd 5422

%.p 241

singlyreinforced section

mkN.M Rd 8037

%.p 472

12

mkN.M Rd 3823

Double reinforced section

mkN.M Rd 1244

• the steel yields both in tension and compression reinforcements

mkN.M Rd 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 x

2

As2 fyd d d2

13

Conclusions:

• in normal reinforcing situations, the compression reinforcement not leads to a significant increaseof the resistance capacity,• to over-reinforced elements, the compression reinforcement leads to a significant increaseof the resistance capacity, • it is not economic to use compression reinforcement in a section before all the singly reinforcingpossibilities are used

14

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 ii

cussi

x

dx j

cu

sj

3

x

dE j

cu

s

sj

13

x

d

x

dE jj

cussj 170013

15

c

j

jsj

t

i

isicd x

dA

x

dhAfxb

1117001700

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

ydsi E

f

ydsi f ifs

ydsi E

f

• for compression reinforcements:

x

dx j

cu

sj

3

x

d.

x

dx jjcusj 1003503 and

ssjsj E ifs

ydsj E

f

ydsj f ifs

ydsj E

f

The resistance moment can be calculated with the following relation:

c

jsjsjsj

t

isisisiccd

c

jsjsj

t

isisiccRd zAzAzfxbzFzFzFM

1111

16

Flanged Section in Bending

• T-sections and L-sections, having their flanges in compression, can both be designed or analyzed in a similar manner,• the flanges generally provide a large compressive area.

FLANGED SECTIONS

Double Reinforced

Singly Reinforced

fhx

fhx

fhx

and lims 1

and lims 1

17

Singly Reinforced Flanged Section –the Depth of the Stress Block Lies Within the Flange:

• , the section can be considered as an equivalent rectangular section of breadth equal to the flange width b. • the non-rectangular section below the neutral axis is in tension and is, therefore, considered to be cracked and inactive

fhx

fhx

18

2f

cdfplateEdh

dfhbMM cdw fdb

21

d

h.

d

h

b

b

fdb

M ff

wcdw

Ed 5012

d

h.

d

h

b

b ff

w501We conclude that: if then the depth of the stress block lies within

the flange.

fhx

The required reinfocement quantity is: zf

MA

yd

Eds 1

• for , we obtain the resistance moment of the plate: MRd=Mplatefhx

the T section is designedas a equivalent rectangularsection h x b :

Singly Reinforced Flanged Section –

fhx

19

fhx Singly Reinforced Flanged Section –

• if , then:fhx

d

h.

d

h

b

b

fdb

M ff

wcdw

Ed 5012

21 ,Rd,RdEd MMM

21111 ,s,ss AAA

21x

dz

22fh

dz

2111

xdfxbzFM cdwc,Rd

2222f

cdfwc,Rdh

dfhbbzFM

and lims 1

20

fhx Singly Reinforced Flanged Section –

221f

cdfwEd,RdEd,Rdh

dfhbbMMMM cdw fdb

21

cdw

fcdfw

cdw

Ed

cdw

,Rd

fdb

hdfhbb

fdb

M

fdb

M

2221 2

d

h.

b

b

d

h f

w

fs 50111

and lims 1

Double Reinforced Section

Singly Reinforced Sectionlims 1

lims 1

21

• we check if the tension steel yields:

lims 1• if , then the flanged section is simply reinforced:

0111 ,c,s FF yd,scdw fAfxb 11

yd

cdw

yd

cdw,s f

fdb

f

fdb

d

xA

11

0221 ,c,s FF yd,scdfw fAfhbb 21

yd

cdfw,s f

fhbbA 21

yd

cdfw

yd

cdw,s,ss f

fhbb

f

fdbAAA 21111

yd

cdw

f

ws f

fdb

d

h

b

bA

11

x

xd

cu

s

3

1

s

ydydcucus E

f.

1

003501

11

331

22

fhx Double Reinforced Flanged Section – and lims 1

321 ,Rd,Rd,RdEd MMMM

3121111 ,s,s,ss AAAA

21x

dz

22fh

dz

23 ddz

2111

xdfxbzFM cdwc,Rd

2222f

cdfwc,Rdh

dfhbbzFM

22323313 ddfAzFzFM ydss,s,Rd

23

• if , then the flanged section is double reinforced, and :lims 1

cdwlim,Rd fdbM 21

22f

cdfw,Rdh

dfhbbM

213 ,Rd,RdEd,Rd MMMM

fcdfwcdwlimEdyds h.dfhbbfdbMddfA 50222

cdw fdb

21

cdw

fcdfw

cdw

cdwlim

cdw

Ed

cdw

yds

fdb

h.dfhbb

fdb

fdb

fdb

M

fdb

ddfA

22

2

2222 50

d

h.

d

h

b

b

fdb

ddfA ff

wlim

cdw

yds 50112

22

yd

cdw

ff

wlim

s f

fdb

dd

d

h.

d

h

bb

A

2

22

5011

ydsyd,s fAfA 231

231 s,s AA

24

cdlimwyd,s fdbfA 11 yd

cdlimw,s f

fdbA 11

cdfwyd,s fhbbfA 21 yd

cdfw,s f

fhbbA 21

3121111 ,s,s,ss AAAA

21 syd

cdfw

yd

cdlimws A

f

fhbb

f

fdbA

21 sfwlimwyd

cds Ahbbdb

f

fA

25

yd

cdw

f

ws f

fdb

d

h

b

bA

11

26

Effective width of flanges

The effective flange width of T beams, depends on:

• the web and flange dimensions,• the type of loading,• the span, • the support conditions,• the transverse reinforcement.

The design of the effective flange width is based on the distance l0 between points of zero moment if:

1,effl 2,effl 3,effl

10 850 ,effl.l

210 150 ,eff,eff ll.l

20 70 ,effl.l

320 150 ,eff,eff ll.l

Section 1-1

23321

/.../l

l

i,eff

i,eff

23 50 ,eff,eff l.l

27