Lec07 Analysis and Design of Doubly Reinforced Beam (Reinforced Concrete Design I & Prof. Abdelhamid Charif)

03-Mar-13
CE 370 : Prof. Abdelhamid Charif 1
CE 370
REINFORCED CONCRETE-I
Prof. Abdelhamid Charif
Analysis and Design of
Doubly Reinforced Beams
Definitions
• The steel that is occasionally used on the compression side of
beams is called compression steel.
• Beams with both tensile and compressive steel are also
referred to as doubly reinforced beams.
• Depth of compression steel is d’ = cover + ds + db’/2
2
d
b
sA
N.A.
'
d
'
sA
c
CE370 : Prof. Abdelhamid Charif
• Bending contribution of steel increases
with lever arm
• For tension steel a high value of depth d
is desirable and for compression steel a
low value of depth d’ is desirable.

03-Mar-13
Why use compression steel ?
• Compression steel is not normally required in beams because
high compressive strength of concrete decreases the need for
such reinforcement.
• Occasionally, however, space or aesthetic requirements limit
beams to such sizes that compression steel is needed in addition
to tensile steel.
• A beam designed with tension steel only but not satisfying
tension-control condition (εt < 0.005) must either have
compression reinforcement or increased dimensions.
• Compression steel may also be provided for practical reasons only
(to fix stirrups and prevent their movement during casting and
vibration)
3CE370 : Prof. Abdelhamid Charif
Why use compression steel ? (cont.)
• Compressive steel increases moment capacity of RC sections as
well as their ductility (higher ultimate curvature).
• Though expensive, compression steel makes beams tough and
ductile, enabling them to withstand large moments,
deformations, and stress reversals such as might occur during
earthquakes.
• Many codes for earthquake zones require that certain minimum
amounts of compression steel be included in flexural members.
• Compression steel reduces long-term deflections due to
shrinkage and plastic flow.
• Tests of doubly reinforced concrete beams have shown that
even if the compression concrete crushes, the beam may very
well not collapse if the compression steel is enclosed by stirrups.

03-Mar-13
5
Effect of compression steel
on sustained load deflection
bd
A'
s
ratiosteelnCompressio' 
6
Effect of compression steel on ductility
Limited effect on strength
Sometimes the neutral axis is close to the compression steel, the
strain and stress are therefore very small. Thus compression
steel adds little moment capacity to the beam. It can, however,
increase beam ductility and reduce long term deflections.

03-Mar-13
CE 370
Analysis of RC Beams with
Compression Steel
Analysis of RC beams
with compression steel
• In beams with compression steel (doubly reinforced),
the amount of tension steel is high (almost maximum).
There is normally no need to check minimum steel
• Flanges in T-beams provide extra compression
capacity. T-beams do not usually require compression
steel
• Analysis and design of doubly reinforced beams is
therefore limited to rectangular sections.

03-Mar-13
Strain Compatibility in
Tension steel in one layer or more.
Compression steel in one layer.
Tensile steel assumed yielding but in
analysis problems, this may be untrue
Compression steel may yield or not.
Strain expressions are derived from
strain compatibility of plane
sections (use of similar triangles).
9
d
b
sA
N.A.
003.0
t
'
s'd
'
sA
c
td
s
mind
min
• As , d , εs = Area, depth and strain at tension steel centroid
• dt , εt = Depth and strain at bottom tension layer (max. depth)
• dmin , εmin = Depth and strain at minimum depth tension layer
• A’s , d’ , ε’s = Area, depth and strain at compression steel
Strain Compatibility in
10
Strain expressions are derived
using similar triangles :















ysy
ysss
s
s
t
t
s
f
E
f
c
dc
c
cd
c
cd
c
cd






'
''
'
'
min
min
if
if
:stresssteelnCompressio
'
003.0:nCompressio
003.0:iondepth tensMin.
003.0:tensionBottom
003.0:tensionCentroid
d
b
sA
N.A.
003.0
t
'
s'd
'
sA
c
td
s
mind
min

03-Mar-13
11
(a) Section
a
abfC cc
'
85.0
'
85.0 cf
ys fAT 
(*)'''
sss fAC 
d
b
sA
N.A.
003.0
'
s'd
'
sA
c
s
(b) Strains (c) Forces
Forces in Doubly Reinforced Beams
 '''
''
'
85.0:concretedisplacedforaccountTo
concrete)displacedconsider(modify to:forcencompressioSteel
85.0:forcencompressioConcrete
assumed)yielding(tensile:forcetensionSteel
csss
sss
cc
ys
ffAC
fAC
abfC
fAT




(*) Compression steel
area displaces same
area of concrete
Displaced Concrete
• Compression steel bars displace an equal amount (area) of
compression concrete.
• The concrete compression force is therefore slightly
reduced
12
 '''
''''''
''
85.0
85.0
:forceecompressivsteelthecorrectingbyaccountintoit
taketoconvenientmorethereforeisItarm.leversametheehasand
levelsteelncompressioat thelocatedisconcretedisplacedThis
85.0reductionForce
csss
scssssss
sc
ffAC
AfAfCAfC
Af




03-Mar-13
13
(a) Section
a
abfC cc
'
85.0
'
85.0 cf
ys fAT 
 ''''
85.0 csss ffAC 
d
b
sA
N.A.
003.0
'
s'd
'
sA
c
s
(b) Strains (c) Forces
Moment in Doubly Reinforced Beams
 
  '''''
''''
85.0
2
85.0
steelileabout tensforcesncompressioofMomentmomentNominal
85.085.0:mequilibriuForce
ddffA
a
dabfM
ffAabffACCT
csscn
csscyssc



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




14
Nominal Moment in
   eq.(b)'''''
85.0
2
85.0momentNominal ddffA
a
dabfM csscn 




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 
  eq.(a)
bf
ffAfA
a
ffAabffACCT
c
csys
csscyssc
s
'
'''
''''
85.0
85.0
:thusisdepthblockStress
85.085.0:mequilibriuForce










ysy
ysss
s
f
E
f


'
''
'
if
if
:stresssteelnCompressio
notoryieldedhassteelncompressiowhetherisdifficultymainThe

03-Mar-13
Analysis of Doubly Reinforced Beams
• Assume tension steel has yielded
• Assume compression steel has yielded
• Compute stress block depth a and neutral axis depth c
• Compute compression and tension steel strains
• If compression and tension steel have yielded, OK continue,
perform strain checks and compute nominal moment
• If compression steel has not yielded, express unknown stress
block depth and compression steel stress in terms of neutral
axis depth c and solve resulting quadratic equation. Perform
strain checks and compute nominal moment.
• If tension steel has not yielded, use adapted method
(Compression steel has yielded)
16
   
c
cd
c
dc
a
c
bf
ffAfA
bf
ffAfA
a
fff
ss
c
cysys
c
csys
yss
s










003.0,003.0:strainsSteel
85.0
85.0
85.0
85.0
:eq.(a)fromdepthblockstressCompute
:yieldedhavesteeltensionandncompressioAssume
'
'
1
'
''
'
'''
'


  ''''
85.0
2
85.0
:eq.(b)hmoment witnominalcompute
:thenassumed,as,yieldedhavesteeltensionandncompressioIf
ddffA
a
dabfM cyscn 







03-Mar-13
(Compression steel has not yielded) (1)
17
 ''''
''
''
'''
85.085.0
:equationmequilibriuforceBack to
600003.0200000
:yieldednothassteelncompressioIf
csscyssc
ysss
sssys
ffAabffACCT
f
c
dc
c
dc
Ef
Ef

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


085.060085.0
85.060085.0
''
'
'
1
'
'
'
'
1
'












ysscsc
cscys
fAAf
c
dc
Acbf
f
c
dc
AcbffA


Substituting the stress block depth (a = β1c) and the compression
steel stress in terms of neutral axis depth gives:
(Compression steel has not yielded) (2)
Multiplying all terms by c and assembling leads to a quadratic
equation with respect to the neutral axis depth c :
 
 
 
   eq.(b)'''''
2''
''''
1
'
'
1
'
'
'
1
'
''''2
85.0
2
85.0:OKIf
yieldncompressionoandyieldioncheck tensandstrainssteelCompute
1
4
1
2
:issolutionpositiveThe
85.0
600
85.0
with
0
ddffA
a
dabfM
RPP
dPRPP
c
b
A
R
bf
A
P
bf
fA
P
dPcRPPc
csscn
s
c
s
c
ys
























03-Mar-13
19
Analysis Problem-1
MPa20andMPa420
shownbeamtheofcapacitymomentdesigntheDetermine
 '
cy ff
All dimensions in mm
750
350
60'
d
202
324
624
66
684d
mm
d
dd
mm
d
dhd
b
s
b
s
60
2
20
1004
2
cover'
68466750
2
32
1004750
2
cover
'














2
2
'
2
2
3.628
4
20
2
0.3217
4
32
4
mmA
mmA
s
s








20
ysys ff  ''
:yieldedhassteelncompressioAssume 
 
  mma
bf
ffAfA
a
c
cysys
527.184
3502085.0
2085.04203.6284200.3217
85.0
85.0
:depthblockstressCompute '
''






yieldingassumedOK0021.000217.0
09.217
6009.217
003.0003.0:strainsteelnCompressio
09.217
85.0
527.184
:depthN.A.
''
'
'
1






yss
s
c
dc
mm
a
c



Solution 1
control-OK tension005.000645.0
09.217
09.217684
003.0003.0:strainTension





tt
t
t
c
cd



03-Mar-13
21
  
  
mkNM
mkNmmNM
M
ddffA
a
dabfM
n
n
n
csscn
.9.7267.80790.0momentDesign
.7.807.107.807
606842085.04203.628
2
527.184
684350527.1842085.0
85.0
2
85.0
:eq.(b)usingmomentNominal
6
'''''


















Solution 1 – Cont.
22
Analysis Problem-2
Analysis of a rectangular section 300 x 600 mm with six 20-mm bars
in two layers as tension steel and three 20-mm bars as compression
reinforcement. Net spacing between steel layers is 30 mm.
Stirrups have 10-mm diameter
MPafMPaf yc 42025'

2'
2
21
2min
12
1
48.942
96.1884
515
2
490
4905054020
54060
60101040'
mmA
mmA
mm
dd
d
mmdd
mmSdd
mmhdd
mmd
s
s
l
t









d
300
206
'd
203
td 600mind

03-Mar-13
23
:yieldedhassteelncompressioAssume 
 
  mma
bf
ffAfA
a
c
cysys
09.62
3002585.0
2585.042048.94242096.1884
85.0
85.0
:depthblockstressCompute '
''






yieldingNot0021.0000536.0
05.73
6005.73
05.73
85.0
09.62
:depthN.A.
''
'
'
1






yss
s
c
dc
mm
a
c



Solution 2
(Compression steel has not yielded)
Neutral axis depth is the solution of a quadratic equation :
 
 
 
 
mmc
c
P
RP
b
A
R
bf
A
P
bf
fA
P
RPP
dPRPP
c
s
c
s
c
ys
046.105
1
696.335765.1041007.146
6035765.1044
1
2
696.335765.1041007.146
35765.104
85.03002585.0
48.942600
696.3
85.0300
48.942
1007.146
85.03002585.0
42096.1884
85.0
600
85.0
with
1
4
1
2
2
'
'
1
'
'
1
'
'
'
1
'
2''
''''









































03-Mar-13
(Compression steel has not yielded old)
25
control-OK tensionandyieldOK005.0and
011.0
046.105
046.105490
003.0
003.0:depthminimumatstrainTension
)420(293.257
046.105
60046.105
600600
yieldingnotOK0021.000129.0
046.105
60046.105
mm105.046c
minmin
min
min
min
'
'
s
''
'
'






















y
y
yss
s
c
cd
MPafMPa
c
dc
f
c
dc
(Compression steel has not yielded)
26
  
  
mkNM
mkNmmNM
M
mmcaMPaf
ddffA
a
dabfM
n
n
n
csscn
.06.33295.36890.0:momentDesign
.95.368.1095.368
605152585.0293.25748.942
2
289.89
515300289.892585.0
289.89046.10585.0and293.257:with
85.0
2
85.0:eq.(b)frommomentNominal
6
1
'
s
'''''





















03-Mar-13
27
Observations about effect of compression steel :
• Three compression steel bars (50% of tension steel) have added
9.36 kN.m (2.9%) only to the beam flexural capacity, because of
the reduced stress (61 % of yield value).
• Little advantage in strength from compression steel
• Main advantage of compression steel is in increasing ductility
and reducing long term deflections
mkNM
mkNM
n
n
.06.33295.36890.0:momentDesign
.95.368:momentNominal



Discussion – Effect of compression steel
mkNM
mkNM
n
n
.70.322559.35890.0:momentDesign
.559.358:momentNominal
:steelncompressiowithoutbeamSame



28
Analysis Problem-3
Analysis of a rectangular section 250 x 600 mm with six 28-mm bars
in two layers as tension steel and two 20-mm bars as compression
reinforcement. Net spacing between steel layers is 28 mm.
Stirrups have 10-mm diameter
MPafMPaf yc 42018'

2'
2
21
12min
1
3.628202
5.3694286
508
2
4805653628
536141040
60101040'
mmA
mmA
mm
dd
d
mmSddd
mmhdd
mmd
s
s
l
t








d
250
286
'd
202
td 600mind

03-Mar-13
29
 
 
yielding)notlayersboth(yieldingnotsteelTension00103.0
05.399
05.399536
003.0003.0:strainsteelTension
yieldingncompressioOK0021.000255.0
05.399
6005.399
05.399
85.0
19.339
:depthN.A.
19.339
2501885.0
1885.04203.6284205.3694
85.0
85.0
:depthblockstressCompute
:yieldedhassteelncompressioAssume
1
1
''
'
'
1
'
''
''


















yt
t
yss
s
c
cysys
ysys
c
cd
c
dc
mm
a
c
mma
bf
ffAfA
a
ff






Solution 3
Tension steel has not yielded
As shown in the previous problem 3, tension steel does not yield
when the neutral axis depth is rather large, which usually leads to
yielding of the compression steel. We therefore assume initially that
compression steel has yielded.
d
b
1
2
s
s
A
A
'd
'
sA
1d2d
Lumping of tension layers cannot be
used since there is no yielding.
We treat the case of problem 3 with
two tension layers in a general way
using strain compatibility.
Both tension layers are not yielding.
2
21 25.1847283 mmAA ss 

03-Mar-13
CE370 : Prof. Abdelhamid Charif 31
yss
ssss
ss
ff
c
dc
c
cd
f
c
cd
Ef
c
cd
c
cd







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''
2
2
1
11
2
2
1
1
'
003.0:stressstrain /nCompressio
600600:stressesTension
003.0003.0:strainsTension



Tension steel has not yielded – Cont.
a
abfC cc
'
85.0
'
85.0 cf
111 ss fAT 
 '''
85.0 cyss ffAC 
2d
b
1
2
s
s
A
A
N.A.
003.0
'
s'd
'
sA
c
1
2
s
s


1d
222 ss fAT 
c
cd
A
c
cd
AffAcbf
TTCC
h
dT
h
dTd
h
C
ah
CM
sscysc
sc
scn

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2
2
1
1
''
1
'
21
2211
600600)85.0(85.0
:mequilibriuForce
22
'
222
:centroidsectionaboutcomputedmomentNominal

a
abfC cc
'
85.0
'
85.0 cf
111 ss fAT 
 '''
85.0 cyss ffAC 
2d
b
1
2
s
s
A
A
N.A.
003.0
'
s'd
'
sA
c
1
2
s
s


1d
222 ss fAT 

03-Mar-13
 
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


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1
)(
)(4
1
2
85.0
)85.0(
85.0
600
85.0
600
0)(
:asexpressed-rebecanhichequation wQuadratic
0)(600)(600)85.0(85.0
600600)85.0(85.0
2
21
221121
1
'
''
1
'
2
2
1
'
1
1
221121
2
221121
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1
'
2
2
1
1
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1
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QPP
dPdPQPP
c
bf
ffA
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bf
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bf
A
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dPdPcQPPc
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c
cd
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c
cd
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c
cys
c
s
c
s
sssscysc
sscysc

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mmc
c
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dPdPQPP
c
bf
ffA
Q
bf
A
bf
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c
cys
c
s
c
s
54.320
1
)21.789.3409.340(
)4809.3405369.340(4
1
2
21.789.3409.340
1
)(
)(4
1
2
21.78
85.02501885.0
)1885.0420(3.628
85.0
)85.0(
9.340
85.02501885.0
25.1847600
85.0
600
85.0
600
:3problemfornapplicatioNumerical
2
2
21
221121
1
'
''
1
'
2
1
'
1
21


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






Tension steel has not yielded – Solution 3 - Cont.

03-Mar-13
kN
c
cd
ATkN
c
cd
AT
kNffACkNcbfC
MPaffMPaEfMPaEf
c
dc
c
cd
c
cd
mmc
ss
cysscc
ysssssss
ys
ys
ys
37.55160001.745600
27.254)85.0(16.104285.0
420,48.298,3.403
OK00244.0
54.320
6054.320
003.0
'
003.0
OK0014924.0
54.320
54.320480
003.0003.0
OK0020165.0
54.320
54.320536
003.0003.0
:forcesandstressesstrains,variousCompute54.320
2
22
1
11
''
1
'
'
2211
'
2
2
1
1
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mkNM
mkNM
h
dT
h
dTd
h
C
ah
CM
n
n
scn
.40.32978.50665.0
yielding)(No65.0
.78.506:obtainon wesubstitutiAfter
22
'
222
:momentNominal 2211



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a
16.1042cC
'
85.0 cf
01.7451 T
27.254sC
2d
b
1
2
s
s
A
A
N.A.
003.0
'
s'd
'
sA
c
1
2
s
s


1d
37.5512 T

03-Mar-13
CE 370
Design of doubly reinforced beams
according to ACI and SBC 304
RC design of beams with compression steel
• If a section, designed with tension steel only, is not tension-
controlled, compression steel is then required.
• The solution is to set tension steel strain equal to minimum
value of 0.005 and then design accordingly.
• The neutral axis depth and strains become therefore all known.
• With many tension layers, satisfying tension-control at the
bottom layer will in general satisfy yield condition at the
tension steel layers.
• If however it is suspected that this might not be true, then the
neutral axis depth will be chosen to satisfy both conditions.

03-Mar-13
39







y
yy
d
cc


10003
3
:depthminimumsteelatYield
min
min




 t
t
tt d
d
cc 375.0
8
3
005.0
:layerbottomatcontrol-Tension











y
t dd
c
10003
3
,
8
3
Min
:conditionsbothSatisfying
min
d
b
sA
N.A.
003.0
005.0t
'
s
'
sA
c
td
s
mind
y min
40
Design steps
cβa
dd
c
y
t
1
min
knownthusisblockstressecompressivconcreteofDepth
steeltensilemaximumProviding
10003
3
,
8
3
Min
conditionsstrainsteelnsionsatisfy tetoaxisneutralSet the2/
layersofnumberexpectedtoaccordingdepthsvariousEstimate1/













continuerequired,steelnCompressioIf
onlysteelh tensionDesign wit
required,notsteelnCompressioIf
2
85.0
:steelncompressiohoutmoment witnominalCompute3/
0
0
'
0









un
un
cn
MM
MM
a
dabfM



03-Mar-13
41
Design steps – Cont.
  
  '''
'
'
'''''
85.0
2
85.0
:areasteelncompressioRequired
85.0
2
85.0
settingbyareasteelncompressiorequiredDetermine5/
ddff
a
dabf
M
A
ddffA
a
dabf
M
M
MM
cs
c
u
s
cssc
u
n
un




















MPaEf
ff
c
dc
ssssys
ysys
s
''''
''
'
200000If
If
'
003.0
:stressandstrainsteelnCompressio4/







42
Design steps – Cont.
 
 
y
cssc
s
csscyssc
f
ffAabf
A
ffAabffACCT
''''
''''
85.085.0
:areasteeltensionRequired
85.085.0
mequilibriuforceusingareasteeltensionrequiredDetermine6/



Although the tension control condition is enforced from the start,
the actually provided steel areas (when choosing bar numbers) may
violate this condition and strain checks are therefore required.
The provided tension steel area must not be excessive as compared
to the required value. This topic will be elaborated later.

03-Mar-13
43
Design Problem-1
85.0MPa30andMPa420
kN.m510andkN.m753
1 

'
cy
LD
ff
MM
mmd
mmdd
mmdd
mmhd
t
70'
67030
73030
700100
min




800
375
'd
'
sA
sA
d
Expecting two tension steel layers, the
various steel depths are estimated as :
Design the shown 375 x 800 mm rectangular beam subjected to
the following dead and live moments :
kN.mM
MMM
u
LDu
0.1392
)510(7.1)375(4.17.14.1:momentUltimate


44
Solution 1
 
mmca
mmc
dd
c
y
t
6875.23275.27385.0
75.27312.394,75.273Min
0021.010003
6703
,
8
7303
Min
10003
3
,
8
3
Min
conditionssteelnsionsatisfy tetoaxisneutralSet the
1
min























requiredsteelnCompressio).1392(
.8.11687.12989.0
.7.1298.107.1298
2
6875.232
7003756875.2323085.0
2
85.0
:steelncompressiohoutmoment witNominal
0
0
6
0
'
0
















mkNMM
mkNM
mkNmNM
a
dabfM
un
n
n
cn



03-Mar-13
45
MPaff
c
dc
ysys
s
420)0021.0(
00223.0
75.273
7075.273
003.0
'
003.0
:stressandstrainsteelnCompressio
''
'








  
  
)4.1140223(8.997
707003085.0420
2
6875.232
7003756875.2323085.0
90.0
101392
85.0
2
85.0
22'
6
'
'''
'
'
mmmmA
A
ddff
a
dabf
M
A
s
s
cs
c
u
s





















46
 
 
)0.6434328(0.6235
420
3085.04208.9973756875.2323085.0
85.085.0
22
''''
mmmmA
A
f
ffAabf
A
s
s
y
cssc
s





4.1140223/8.997
:)(steelncompressioActual/Required
0.6434328/0.6235
:)(steeltensionActual/Required
'
2
2


s
s
A
mm
A
mm
800
375
'd
223
328
d

03-Mar-13
Using 32 mm for bar spacing (not less than bar diameter), and
10 mm stirrups, the maximum number of 32-mm bars in one layer is:
4
67.4
3232
0423210632375
cover26
max 







n
n
Sd
ddSb
n
bb
bsb
47
Two layers are therefore required as assumed (4+4).
For layer spacing, we use 30 mm.
800
375
'd
223
328
d
375
'd
223
328
d
4.1140223/8.997:)(steelncompressioActual/Required
0.6434328/0.6235:)(steeltensionActual/Required
'2
2


s
s
Amm
Amm
mm
d
dd
mm
dd
d
mmdSddd
mmdd
d
dhdd
b
s
bl
t
b
st
61111040
2
cover'
703
2
6723230734
734161040800
2
cover
'
21
1min2
1
1







The difference between the required and provided values of steel
areas may lead to violation of the tension control condition. Strain
checks are therefore required using actual steel areas and depths .

03-Mar-13
375
'd
223
328
d
mmd
mm
dd
d
mmdd
mmdd t
61'
703
2
672
734
21
min2
1






The actual tension steel depths are slightly greater than was
initially safely assumed.
The actual compression steel depth is less than was assumed and
this also is on the safe side.
The steel strain checks are however still required because of the
excess in the provided steel areas.
50
:yieldedhassteelncompressiothatbeforefoundwasasAssume

 
 
11.277
6111.277
11.277
85.0
54.235
:depthN.A.
54.235
3753085.0
3085.04204.11404200.6434
85.0
85.0
:depthblockstressactualCompute
''
'
'
1
'
''












yss
s
c
cysys
c
dc
mm
a
c
mma
bf
ffAfA
a




03-Mar-13
51
control-tensionOKNOT005.000495.0
00495.0
11.277
11.277734
003.0003.0
:layer)(bottomdepthmaximumatcheckcontrolTension
yieldOK0043.0
11.277
11.277672
003.0003.0
:depthminimumatcheckYield
min
min











t
t
t
y
c
cd
c
cd



The section turns out to be not tension controlled because in the
provided steel areas, there was more excess in the tension steel:
2'
2
6.1428.9974.1140:nCompressio
0.19900.62350.6434:Tension
:Required-ProvidedareasteelExcess
mmA
mmA
s
s



375
'd
282
328
d
2'
2
'2
2
7.2338.9975.1231:nCompressio
0.19900.62350.6434:Tension
5.1231282/8.997:)(steelncompressioActual/Required
0.6434328/0.6235:)(steeltensionActual/Required
mmA
mmA
Amm
Amm
s
s
s
s




mm
d
dd
mmddd
mmddmmdd
b
s
t
64141040
2
cover'
:depthsteelncempressioNew
7032/)(
672734
:beforeasaredepthssteelTension
'
21
min21



Let us now try another solution with more excess in compression
steel (keeping the same bar reinforcement in tension):

03-Mar-13
53
:yieldedhassteelncompressioagain thatAssume 
 
 
689.272
64689.272
689.272
85.0
786.231
:depthN.A.
786.231
3753085.0
3085.04205.12314200.6434
85.0
85.0
:depthblockstressactualCompute
''
'
'
1
'
''












yss
s
c
cysys
c
dc
mm
a
c
mma
bf
ffAfA
a



54
control-tensionOK005.000508.0
689.272
689.272734
003.0
003.0:layer)(bottomdepthmaximumatcheckcontrolTension
yieldOK00439.0
689.272
689.272672
003.0003.0
min
min










t
t
t
y
c
cd
c
cd



  
  
OKmkNMM
mkNmmNM
M
ddffA
a
dabfM
un
n
n
csscn

















.0.13923.148974.16549.0momentDesign
.74.1654.1074.1654
647033085.04205.1231
2
786.231
703375786.2313085.0
85.0
2
85.0:(optional)checkMoment
6
'''''


03-Mar-13
Over Design and Tension Control Condition
• Recall the maximum tensile steel ratio which corresponds to a
tensile strain of 0.005 (Figure a).
• To maintain force equilibrium, any further increase in the
tensile steel area would require a longer compression block
and thus a shift downwards for the neutral axis (Figure b).
• This reduces the steel strain below the limit of 0.005 and
violates the tension control condition of SBC code.
d
areasteelMaximum:(a)
max,sA
003.0
005.0t
maxc
d
steelmax.thanMore:(b)
max,sA
003.0
005.0t
maxc
• In the design of beams with tension steel only, the risk of over
design, generating non-tension-controlled beams, is avoided by
imposing a steel strain check using the actual provided steel
area and depth (equivalent to satisfying maximum steel limit).
• The over design risk is higher in beams requiring compression
steel, as the design is based on the limit condition.
• Excess in provided tensile steel area will also shift the neutral
axis downwards and violate the tension control condition.
d
steelRequired:(a)
sA
003.0
005.0t
maxc
d
in tensionexcessmorewithsteelProvided:(b)
psA ,
003.0
005.0t
maxc'
sA '
, psA

03-Mar-13
• Excess in tensile steel must be balanced by corresponding
excess in compression steel.
• Thus in order to avoid violation of the tension control condition,
the required compression steel area must be re-determined
using the actual tension steel area (after setting bar diameter
and bar number for tension steel).
• The new compression steel area is determined using force
equilibrium and using the actual provided tension steel area:
 
 ''
'
,'
2,
''''
,
85.0
85.0
:isareasteelncompressiorequiredNew
85.085.0
cs
cyps
s
csscypssc
ff
abffA
A
ffAabffACCT




Design Summary (if compression steel required
1. Assume initial steel depths according to expected layers (do not
overestimate tension depth / underestimate compression depth)
2. Set neutral axis depth to satisfy tension steel strain conditions
and deduce compression steel strain and stress.
3. Compute required compression steel using force equilibrium.
4. Deduce required tension steel using moment equilibrium.
5. Select bar diameter, find required bar number for tension steel,
and then determine the actual provided area.
6. Deduce the new required compression steel using force
equilibrium and the actual provided tension steel area.
7. Select bar diameter, find required bar number for compression
steel, and then determine the actual provided area
8. Determine final values of steel depths, and perform moment
check if required.

03-Mar-13
59
  
 
y
cssc
s
cs
c
u
s
ssssys
ysys
s
y
t
f
ffAabf
A
ddff
a
dabf
M
A
MPaEf
ff
c
dc
cβa
dd
c
''''
'''
'
'
''''
''
'
1
min
85.085.0
:areasteeltensionRequired/4
85.0
2
85.0
:areasteelncompressioRequired3/
200000If
If
'
003.0and
10003
3
,
8
3
Min
conditionsstrainsteelnsionsatisfy tetoaxisneutralSet the2/
layersofnumberexpectedtoaccordingdepthsvariousEstimate1/









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
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
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














RC design steps for beams with compression steel
60
RC design steps for beams with compression steel – Cont.
 
design-rethen),(unsafestillifand
checkmomentperformunsafeifdepths,steelfinalDeduce/8
4
withSelect/7
85.0
85.0
:areasteelncompressiorequiredNew6/
4
withSelect/5
'''
,
2'
'
'
'
2,''
''
'
,'
2,
,
2
b
b
un
bbps
b
b
b
s
bb
cs
cyps
s
bbpsb
b
s
b
MM
ANA
d
A
A
A
Nd
ff
abffA
A
ANA
d
A
A
A
Nd









• With this strategy, the tension control condition is
always satisfied.

03-Mar-13
61
Re - Solve Design Problem-1
85.0MPa30andMPa420 1  '
cy ff
mmd
mmdd
mmdd
mmhd
t
70'
67030
73030
700100
min




800
375
'd
'
sA
sA
d
No need to show again that compression
steel is required. We use the same initial
steel depths:
Design the shown 375 x 800 mm rectangular beam subjected to
an ultimate moment of 1392 kN.m:
62
Re-Solution 1
 
mmca
mmc
dd
c
y
t
6875.23275.27385.0
75.27312.394,75.273Min
0021.010003
6703
,
8
7303
Min
10003
3
,
8
3
Min
conditionssteelnsionsatisfy tetoaxisneutralSet the
1
min























MPaff
c
dc
ysys
s
420)0021.0(
00223.0
75.273
7075.273
003.0
'
003.0
''
'









03-Mar-13
63
Re-Solution 1 – Cont.
  
  
2'
6
'
'''
'
'
8.997
707003085.0420
2
6875.232
7003756875.2323085.0
90.0
101392
85.0
2
85.0
mmA
A
ddff
a
dabf
M
A
s
s
cs
c
u
s





















 
 
2
,
2
''''
0.6434Φ3288:barsΦ32Using
0.6235
420
3085.04208.9973756875.2323085.0
85.085.0
mmAN
mmA
f
ffAabf
A
psb
s
y
cssc
s






Re-Solution 1 – Cont.
 
 
2'
,
2'
2,
'
2,
''
'
,'
2,
5.1231282Use65.1200
3085.0420
3756875.2323085.04200.6434
85.0
85.0
:areasteelncompressiorequiredNew
mmAmmA
A
ff
abffA
A
pss
s
cs
cyps
s







375
'd
282
328
d
The required compression steel area
jumped from 997.8 to 1200.65 mm2
because of the provided tension steel.
This time, we find directly the second
solution which was checked before.

03-Mar-13
65
Design Problem-2
0.85MPa30andMPa420 1  '
cy ff
600
375
'd
'
sA
sA
d
Design the shown 375 x 600 mm rectangular beam for an ultimate
moment of 780 kN.m and with the following data:
mmd
mmdd
mmdd
mmhd
t
60'
47030
53030
500100
min




Expecting two tension steel layers, the
various steel depths are estimated as :
66
Solution 2
 
mmca
mmc
dd
c
y
t
9375.16875.19885.0
75.19847.276,75.198Min
0021.010003
4703
,
8
5303
Min
10003
3
,
8
3
Min
1
min























requiredsteelnCompressio).780(
.15.60428.67190.0
.28.671.1028.671
2
9375.168
5003759375.1683085.0
2
85.0
0
0
6
0
'
0
















mkNMM
mkNM
mkNmNM
a
dabfM
un
n
n
cn


Set the N.A. depth to satisfy tension steel conditions:
Nominal moment without compression steel :

03-Mar-13
67
MPaEf
c
dc
sss
ys
s
8.418002094.0200000
yieldedNot)0021.0(
002094.0
75.198
6075.198
003.0
'
003.0
''
'
'










  
  
2'
6
'
'''
'
'
09.1129
605003085.08.418
2
9375.168
5003759375.1683085.0
90.0
10780
85.0
2
85.0
mmA
A
ddff
a
dabf
M
A
s
s
cs
c
u
s





















68
 
 
   
2'
,
2'
2,
''
'
,'
2,
2
,
2
''''
64.1256204Use95.1152
3085.08.418
3759375.1683085.04200.4926
85.0
85.0
:areasteelncompressiorequiredNew
4)(4layersOK two0.4926288:bars28Using
66.4903
420
3085.08.41809.11293759375.1683085.0
85.085.0
mmAmmA
ff
abffA
A
mmA
mmA
A
f
ffAabf
A
pss
cs
cyps
s
ps
s
s
y
cssc
s














03-Mar-13
mm
d
dd
mm
dd
d
mmdSddd
mmdd
d
dhdd
b
s
bl
t
b
st
60101040
2
cover'
507
2
4782830536
536141040600
2
cover
'
21
1min2
1
1







375
'd
204
288
d600
Actual steel depths are :
The actual compression steel depth is unchanged whereas the
tension steel depths are slightly greater than was initially safely
assumed. No moment check is therefore required.
Solution 2 – Strain Check – Optional only
Assume as was found that compression steel has not yielded. The
neutral axis depth is thus the solution of a quadratic equation :
 
 
 
 
mmcammc
c
P
RP
b
A
R
bf
A
P
bf
fA
P
RPP
dPRPP
c
s
c
s
c
ys
20.165357.19485.0357.194
1
9423.37594.925384.254
607594.924
1
2
9423.37594.925384.254
7594.92
85.03753085.0
6.1256600
9423.3
85.0375
6.1256
5384.254
85.03753085.0
4200.4926
85.0
600
85.0
with
1
4
1
2
1
2
'
'
1
'
'
1
'
'
'
1
'
2''
''''










































03-Mar-13
71
)420(8.414002074.0200000
yieldingnotOK0021.0002074.0
357.194
60357.194
''
s
''
'
'
MPafMPaEf
c
dc
yss
yss
s









controlTensionOK005.000527.0
357.194
357.194536
003.0003.0
:layer)(bottomdepthmaximumatcheckcontrolTension
yieldOK0044.0
357.194
357.194478
003.0003.0
min
min
min










t
t
t
y
c
cd
c
cd




72
  
  
surprise)(No.0.780
.196.800105.88990.0:momentDesign
.105.889.10105.889
605073085.08.4146.1256
2
2.165
5073752.1653085.0
85.0
2
85.0
6
'''''
OKmkNMM
mkNM
mkNmmNM
M
ddffA
a
dabfM
un
n
n
n
csscn



















Moment check (not required)
Since the tensile steel strain conditions are satisfied and the
actual steel depths are safer than assumed, the moment check is
not required. We still perform it.

03-Mar-13
Design of doubly reinforced beams
Final observations
• It is important in the final design stage to comply with the design
results. Any excess in the tensile steel area may result in
violation of tension-control condition.
• Excess in tensile steel must be balanced by corresponding excess
in compression steel.
• Excess of steel may increase moment capacity but may change
the type of beam behavior and failure
• Over-design is not only uneconomical, it may also affect
structural safety
• The design method described previously avoids systematically
over design leading to violation of tension control condition.
Effect of tension steel ratio
These moment curvature
relationships, produced by
RC-TOOL software, show
the ductility of under
reinforced beams.
Despite their improved
strength, over reinforced
beams have little energy
absorption capacity, which
is the main factor in seismic
design.

03-Mar-13
Effect of compression steel ratio
These moment curvature
relationships, produced by
RC-TOOL software, show
that adequate compression
steel can recover the
ductility of a brittle beam.
A measure of ductility
(energy absorption
capacity) is given by the
area under the curve.
Maximum compression steel ?
• SBC and ACI codes do not fix any maximum limit to a beam total
steel reinforcement.
• The maximum tension steel ratio is satisfied through the tension
control condition.
• If the tension control check is not satisfied, then either the section
is increased or compression steel must be provided.
• There is a risk that the second option will be abusively used
resulting in excessive compression steel.
• Should not there be a limit on the compression steel ?
• The deficiency in the section may be such that the required
compression steel may be too large and even exceed tension steel.
• This strange situation is more likely in shallow beams which are
used for architectural reasons (popular in KSA).

03-Mar-13
Design Problem 3
Design of a shallow beam (depth less than width) 1000 x 320 mm.
In order to hide the beam, its thickness was limited to that of the
slab (320 mm).
This “hiding” architectural constraint is mechanically obsolete and
must be restrained by the structural community.
We design this beam for four different values of the ultimate
moment (300, 400, 500, 600 kN.m).
1000
320
MPa420MPa20  y
'
c ff
For simplification, we consider
the same steel depths in all
cases based on one layer in both
tension and compression steel.
d = h – 60 = 260 mm, d’ = 60 mm
78
Solution 3
mmca
mm
d
c
875.825.9785.0
5.97
8
2603
8
3
layer)tension(oneconditionssteelnsionsatisfy tetoaxisneutralSet the
1 




cases.allinrequiredsteelnCompressio
.1.2779.30790.0
.9.307.109.307
2
875.82
2601000875.822085.0
2
85.0
:steelncompressiohoutmoment witNominal
0
0
6
0
'
0
















un
n
n
cn
MM
mkNM
mkNmNM
a
dabfM



03-Mar-13
79
ys
y
s
sssys
s
ff
f
f
MPaEf
c
dc
of%5555.0
8.230001154.0200000
001154.0
5.97
605.97
003.0
'
003.0
'
'
'''
'









  
 
y
cssc
s
cs
c
u
s
f
ffAabf
A
ddff
a
dabf
M
A
''''
'''
'
'
85.085.0
85.0
2
85.0












300 594.2 3656.9 0.16 0.013
400 3193.1 4979.7 0.64 0.026
500 5792.0 6302.4 0.92 0.038
600 8390.8 6525.2 1.29 0.047
mkN
Mu
. bh
AA ss
'

2
'
mm
As
2
mm
As
s
s
A
A'
The following table presents the various results of required
tension and compression steel areas for the four ultimate
moment values. Comparison ratios are also given.

03-Mar-13
Maximum compression steel ?
• We can observe the rapid increase in the compression
steel area, until it exceeds the tension steel value.
• To avoid such nonsense situations, a limit must be set.
• Despite the “silence” of SBC and ACI codes on this
issue, it is recommended to limit compression steel
amount to a fraction of tension steel area (say 20 to
40%) or to fix a limit on the total steel ratio.
Analysis of a general multi-layer RC section
using strain compatibility method - 1
• Many problems were previously solved using strain compatibility.
• A general RC section with many steel layers is now considered
with partial yielding. The most strained layers are yielding
whereas the remaining layers are in the elastic stage.
• This allows tuning to any possible case.
• The steel area, depth and strain at each layer are :
• Asi , di , εsi = At
tension layer i
• εt = At bottom layer
(tension control, Ф)
• A’
sj , d’
j , ε’
sj = At
compression layer j
0.003
di
a =
β1c
Depths Strains Forces
c
'
jd
si
t
abfc
'
85.0
'
85.0 cf
'
sj
 '''
85.0 csjsj ffA 
sisi fA

03-Mar-13
• Steel stresses are :
















c
dc
Ef
ff
c
cd
Ef
ff
j
sjssjysj
ysjysj
i
sissiysi
ysiysi
'
'''
''
600:If
:If
600:If
:If




0.003
di
a =
β1c
c
'
jd
si
t
abfc
'
85.0
'
85.0 cf
'
sj
 '''
85.0 csjsj ffA 
sisi fA
c
dc
c
cd
j
sj
i
si
'
'
003.0
003.0






• Steel strains are :
   ''''
85.085.0 csjsjcsisisjcsi ffAabffACCT
• Force equilibrium:
Starting from an initial guess, steel layers are split in yielding (y) and
elastic (e) parts, and using a = β1c, equilibrium equation becomes:
      '''''
1
'
85.0
600
85.0
600
sjcejsejsyjyceiseisyiy AfcdA
c
AfcbfcdA
c
Af 
   ''''
85.085.0 csjsjcsisi ffAabffA
Multiplying by c and grouping leads to a general quadratic equation :






''
0
1
'
'
1
'
'
'
1
'
'
'
1
'
1
'
'''
0
''2
0If:caseSpecial
85.0
600
85.085.0
600
85.0
with0
ejejeiei
sj
c
sej
ej
c
yysj
yj
c
esi
ei
c
ysyi
yi
eiyiejyjejejeieio
dPdPcP
b
A
R
bf
A
P
bf
fA
P
bf
A
P
bf
fA
P
RPPPPPdPdPcPc


03-Mar-13
The positive solution of this quadratic equation is :
1
'
'
1
'
'
'
1
'
'
'
1
'
1
'
'''
0
''2
85.0
600
85.085.0
600
85.0
with0
 b
A
R
bf
A
P
bf
fA
P
bf
A
P
bf
fA
P
RPPPPPdPdPcPc
sj
c
sej
ej
c
yysj
yj
c
esi
ei
c
ysyi
yi
eiyiejyjejejeieio




 











 1
4
1
2 2
0
''
0
P
dPdPP
c
ejejeiei
Depending on the layers and their yielding state, the term P0 may
be positive or negative. The expression of the solution is sign
dependent. The absolute value and (±) are required for generality.
(+) sign is used if the term P0 is negative and (-) is used otherwise.
All strains are deduced
and if the assumed
yielding is violated in
any layer, a new
iteration must be
carried out using the
last yielding state.
  

















  22
85.0
22
85.0 ''''' h
dfAd
h
ffA
ah
abfM isisijcsjsjcn
Bottom tensile layer strain εt is used to check tension control and
obtain the strength reduction factor leading to design moment ФMn
The nominal moment is obtained about the section centroid as :
0.003
di
a =
β1c
'
jd
si
t
abfc
'
85.0
'
85.0 cf
'
sj
 '''
85.0 csjsj ffA 
sisi fA
h/2

03-Mar-13
Example of multi layer general section
Strain compatibility method - 1
Analyze the 600 x 1500 shown
section with the following
material and layer spacing data :
mmS
.εMPaf
.βMPaf
l
yy
'
c
30
00260520
85022 1



mm600
1000
204
2883 
375
1500
208
162
162
There are two skin layers with two Φ16
bars each. Top skin layer (d =375mm)
is considered part of the compression
steel and the second layer (d =1000mm)
is part of tension steel.
mmd
mmdSdd
mm
d
dd
mmd
mmdSdd
mmdSdd
mmd
d
dhdd
bl
b
s
bl
bl
b
st
375
110203060
60101040
2
cover
1000
1320581378
1378581436
14361410401500
2
cover
'
3
''
1
'
2
'
'
1
4
23
12
1
1








mm600
1000
204
2883 
375
1500
208
162
162
There are four tension layers and
three compression layers and the
depths are :

03-Mar-13
2
2
'
3
2
2
'
1
2
2
'
1
2
2
4
2
2
321
1.402
4
16
2
6.1256
4
20
4
2.2513
4
20
8
1.402
4
16
2
0.4926
4
28
8
mmA
mmA
mmA
mmA
mmAAA
s
s
s
s
sss










mm600
1000
204
2883 
375
1500
208
162
162
The various areas are :
We start by assuming that all steel has yielded except the skin
layers which are closer to the neutral axis. All layers in the yield
parts (y) except tension layer 4 and compression layer 3 which are
in the elastic parts (e).
The neutral axis depth is given by :
 
1
'
'
1
'
'
'
1
'
'
'
1
'
1
'
'''
02
0
''
0
85.0
600
85.085.0
600
85.0
with1
4
1
2
 b
A
R
bf
A
P
bf
fA
P
bf
A
P
bf
fA
P
RPPPPP
P
dPdPP
c
sj
c
sej
ej
c
yysj
yj
c
esi
ei
c
ysyi
yi
eiyiejyj
ejejeiei

















03-Mar-13
The various parameters are :
1802.8
85.0600
1.4026.12562.2513
2973.25
85.06002285.0
1.402600
85.0
600
5155.68
85.06002285.0
5206.1256
85.0
0309.137
85.06002285.0
5202.2513
85.0
2973.25
85.06002285.0
1.402600
85.0
600
5876.268
85.06002285.0
5200.4926
85.0
1
'
3
'
2
'
1
1
'
'
11
'
'
3'
3
1
'
'
2'
2
1
'
'
1'
1
11
'
4
4
1
'
1
321

































b
AAA
b
A
R
bf
A
P
bf
fA
P
bf
fA
P
bf
A
P
bf
fA
PPP
ssssj
c
s
e
c
ys
y
c
ys
y
c
s
e
c
ys
yyy
 
   
mmc
dPdP
c
P
dPdPP
cP
P
P
RPPPPPPPRPPPPP
RPPP
PPPP
eeee
ejejeiei
eyyyeyyeiyiejyj
eyy
eyyy
415.614
1
802.557
37510002973.254
1
2
802.557
1
802.557
4
1
2
802.557
1
4
1
2
0
802.557
1802.82973.255876.26832973.255155.680309.137
1802.82973.255155.680309.137
2973.255876.268
22
'
3
'
344
2
0
''
0
0
0
0
'
4321
'
3
'
2
'
1
'''
0
''
3
'
2
'
1
4321










































03-Mar-13
OK:00117.0
415.614
375415.614
003.0
OKNOT:00246.0
415.614
110415.614
003.0
OK:0027.0
415.614
60415.614
003.0
OK:001883.0
415.614
415.6141000
003.0
OK:00345.0
415.614
415.6141320
003.0
OK:0037.0
415.614
415.6141378
003.0
OK:0040.0
415.614
415.6141436
003.0
003.0003.0
'
3
'
2
'
1
4
3
2
1
'
'
ys
ys
ys
ys
ys
ys
ys
j
sj
i
si
c
dc
c
cd

































c = 614.415 mm
Steel layer strains are :
The initial assumed
yielding state is violated
at the compressive layer 2
A second round of
calculations is therefore
required using the last
yielding state.
All tension parameters and some of the compression
parameters remain unchanged. The new values are :
 
  mmc
dPdPdP
c
P
RPPPPPPPRPPPPP
bf
A
P
b
A
RPP
PPPP
eeeeee
eyyyeeyeiyiejyj
c
s
e
sj
ey
eyyy
656.6171
261.547
3752973.251100563.7910002973.254
1
2
261.547
1
261.547
4
1
2
261.547
261.5471802.82973.255876.26832973.250563.790309.137
0563.79
85.06002285.0
6.1256600
85.0
600
1802.82973.250309.137
2973.255876.268
2
2
'
3
'
3
'
2
'
244
0
'
4321
'
3
'
2
'
1
'''
0
1
'
'
2'
2
1
'
''
3
'
1
4321



































03-Mar-13
OK:0011786.0
656.617
375656.617
003.0
OK:0024657.0
656.617
110656.617
003.0
OK:00271.0
656.617
60656.617
003.0
OK:001857.0
656.617
656.6171000
003.0
OK:00341.0
656.617
656.6171320
003.0
OK:00369.0
656.617
656.6171378
003.0
OK:003975.0
656.617
656.6171436
003.0
003.0003.0
'
3
'
2
'
1
4
3
2
1
'
'
ys
ys
ys
ys
ys
ys
ys
j
sj
i
si
c
dc
c
cd

































c = 617.656 mm
Steel layer strains are :
The assumed yielding
state is now confirmed in
all layers.
The stresses are :
ssssys
ysys
Ef
MPaff




200000
520
Compute the various forces to check equilibrium :
 
 
 
OKkNT
NfAT
NfATTT
NffAC
NffAC
NffAC
NabfC
MPafmmca
sss
yssss
csss
csss
csss
cc
c








 9.7833C:checkmequilibriuForce
0.1493404.3711.402
0.25615205204926
7.87255)7.187.235(1.40285.0
3.596181)7.1814.493(6.125685.0
2.1259867)7.18520(2.251385.0
2.589061260001.5252285.085.0
7.1885.001.525656.61785.0
444
1321
''
3
'
32
''
2
'
22
''
1
'
11
'
'
1

03-Mar-13
The nominal moment is obtained about the section centroid as :
 




























































































2
1500
10004.3711.402
2
1500
13205204926
2
1500
13785204926
2
1500
14365204926
375
2
1500
)7.187.235(1.402110
2
1500
)7.1814.493(6.1256
60
2
1500
)7.18520(2.2513
2
01.525
2
1500
60001.5252585.0
22
85.0
22
85.0
2222
'''''
'
n
isisijcsjsjcn
isijsjcn
M
h
dfAd
h
ffA
ah
abfM
h
dTd
h
C
ah
CM
The nominal moment is :
The bottom tensile layer strain εt is used to check tension control
and obtain the strength reduction factor.
mkNMn
y
yt
st
.45.715347.90187932.0
7932.0
0026.0005.0
0026.0003975.0
0.250.65
005.0
0.250.65
ionIn transit003975.01














mkNmmNMn .47.9018.1047.9018 6


Lec07 Analysis and Design of Doubly Reinforced Beam (Reinforced Concrete Design I & Prof. Abdelhamid Charif)

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