Lec04 Analysis of Rectangular RC Beams (Reinforced Concrete Design I & Prof. Abdelhamid Charif)

25-Feb-13
CE 370 : Prof. Abdelhamid Charif 1
CE370
REINFORCED CONCRETE-I
Prof. Abdelhamid CHARIF
Ultimate Flexural Analysis of Beams
According to SBC / ACI Codes
Concrete Stress-Strain
2
• Ultimate concrete
strain varies between
0.003 and 0.0045
• Lower strength
concrete has a higher
value .
• SBC and ACI codes fix
the ultimate concrete
compressive strain to
0.003 for all grades of
normal concrete.
003.0cu

25-Feb-13
Stress Distribution at Ultimate Stage
3
d
b
sA
(a) Section
(b) Strain
(d) Whitney’s
assumed stress
N.A.
003.0
s (c) Actual
stress
ca 1
abfC c
'
85.0
'
85.0 cf
ss fAT 
c
sf
'
cf
Whitney replaced the curved stress block with an equivalent
rectangular one having the same area and the same centroid.
 65.0,008.009.1Max:MPa30For
85.0:MPa30For
depthand850ofstressConstant
1
1
1
'
c
'
c
'
c
'
c
ff
f
caf.






Nominal and Design Flexural Strength
4
d
b
sA
N.A.
c ca 1 abfC c
'
85.0
'
85.0 cf
ss fAT 
un MM 
Theoretical or Nominal Flexural Moment (Strength) noted : Mn
Design or Usable Flexural Moment (Strength) noted : Mn
Design Moment = Nominal Moment X Strength Reduction Factor
Design moment must be equal to or greater than ultimate moment
caused by factored loads:

25-Feb-13
5
• In all structural members, moments are computed about the
centroid of the gross section.
• For beams with no external force, the two internal forces are
equal C = T. Thus : Moment = Moment of a force couple
• Moment = T x Lever arm = C x Lever arm.
Moments in RC Beams
d
b
sA
N.A.
c ca 1 abfC c
'
85.0
'
85.0 cf
ss fAT 













22
a
dC
a
dTMn
Tension steel ratio for RC beams
• In reinforced concrete beams subjected to bending,
it is common to express the tension steel amount
using either the area As or the corresponding steel
ratio (no unit) defined as:
6
bdA
bd
A
s
s
 
Where b and d are the width of the section and the
depth of the tension steel respectively.
Note that the steel ratio in beams is not related to the
gross section area as in columns.

25-Feb-13
Minimum Steel Ratio
• The applied bending moment (Mu) may sometimes be so small
that an unreinforced concrete section can theoretically resist it.
• However, the section will fail immediately when a crack occurs.
• The cracks may occur even if the member is unloaded because of
shrinkage effects. This type of failure is brittle with no warning.
• To prevent such a possibility, codes specify a minimum amount of
reinforcement that must be satisfied at every section of flexural
members.
7



















yy
c
w
s
w
w
yy
c
s
ff
f
db
A
b
db
ff
f
A
4.1
,
4
Max
beamofwidthwebwhere
4.1
,
4
Max
'
min
min,
min
'
min,

Section Types Based on Ductility
8
memberscontrollednCompressioorBrittleCalled
yieldssteelbeforecrushesConcrete
:)reinforced-(overfailureBrittle
memberscontrolledTensionorDuctileCalled
crushesconcretebeforeyieldsSteel
:)reinforced-(underfailureDuctile
members/sectionsBalancedCalled
crushesconcretewhenyieldreachesSteel
:sectionBalanced






b
b
b




25-Feb-13
Balanced Section
 A section that has a steel ratio such that the steel
reaches yield strain εy (= fy /Es) when the concrete
strain attains the ultimate value of 0.003.
9
d
b
sA
c
003.0
cd 
syy Ef /
b
s
bd
A




:sectionbalancedFor
ratioStee
Brittle Members
 Members whose steel tensile strain εt at the extreme bottom
layer is equal to or less than εy when the concrete strain
reaches 0.003 are called compression-controlled –
considered to be brittle by SBC and ACI.
 Concrete crushes before steel yields.
 Deflections are small and there is little warning of failure.
10
d
b
sA
c
003.0
cd 
yt  
ts
t
s
εε
ε
ε
:onlylayeroneofIn
layerbottomextremeatStrain:
centroidsteelat tensionStrain:
:usedNotations

25-Feb-13
Ductile members
 Members whose steel tensile strain εt at the extreme bottom
layer is equal to greater than 0.005 (more than yield strain)
when the concrete strain reaches 0.003 are called tension-
controlled – considered to be fully ductile by SBC and ACI
 Steel yields well before concrete crushes
 Deflections are large and there is warning before failure
11
d
b
sA
c
003.0
cd 
005.0t
12
Transition Region
 Members with steel strains between y and 0.005 are
in a transition region (y < εt < 0.005)
12
d
b
sA
c
003.0
cd 
005.0 ty 

25-Feb-13
Balanced Steel Ratio
13
d
b
sA
N.A.
003.0
yt  
ca 1 abfC c
'
85.0
'
85.0 cf
ys fAT 
c
At the balanced point, steel reaches its yield strain at the same
time as concrete reaches its ultimate strain of 0.003
The corresponding steel ratio is called balanced steel ratio ρb
The balanced steel ratio is obtained by equating two different
expressions of the neutral axis depth.
14
(b)










s
y
y
E
f
d
c
d
c
003.0
003.0
003.0
003.0
:ianglessimilar trFrom

(a)'
c
y
s
'
c
y
yys
'
c
f
df
c
a
c
bd
A
f
df
a
bdffAabf.
TC
11 85.0
with,
85.0
850
:mequilibriuForce










d
b
sA
003.0
s
y
y
E
f

ca 1
abfC c
'
85.0
'
85.0 cf
ys fAT 
c

25-Feb-13
15

















































yy
'
c
yy
'
c
b
yy
'
c
yb
s
s
y
'
c
yb
ff
f
f
f
fff
f
E
E
f
d
f
df
600
60085.0
003.0
003.085.0
600
600
2000000
003.0
003.0
85.0
MPa)102(
003.0
003.0
85.0
11
1
6
1








d
b
sA
003.0
s
y
y
E
f

ca 1
abfC c
'
85.0
'
85.0 cf
ys fAT 
c
(b)
(a)









s
y
'
c
y
E
f
d
c
f
df
c
003.0
003.0
85.0 1

Maximum Steel Ratio
16
 
(b)dc
d
c
8
3
8
3
005.0003.0
003.0
:ianglessimilar trFrom



d
b
sA
003.0
005.0
ca 1
abfC c
'
85.0
'
85.0 cf
ys fAT 
c
In order to have the members ductile enough, steel tensile strain
should not be less than 0.005 (when concrete strain reaches 0.003)..
This corresponds to the maximum tension steel ratio. It is again
obtained by equating two expressions of the neutral axis depth.
(a)'
c
y
s
'
c
y
yys
'
c
f
df
c
a
c
bd
A
f
df
a
bdffAabf.
TC
11 85.0
with,
85.0
850
:mequilibriuForce











25-Feb-13
Maximum Steel Ratio
17
max
1
max
1
max
1
max
005.0
8
385.0
8
385.0
8
3
85.0










t
y
'
c
y
'
c
'
c
y
ε
f
f
f
fd
f
df
d
b
sA
003.0
005.0
ca 1
abfC c
'
85.0
'
85.0 cf
ys fAT 
c
(b)
(a)
dc
f
df
c '
c
y
8
3
85.0 1




18
b
y
y
b
y
'
c
yy
'
c
b
f
f
f
f











008.0
003.0
3
8
003.0
003.0
8
385.0
003.0
003.085.0
:ratiostwofor thesexpressionpreviousUsing
max
max
1
max
1








d
003.0
max
005.0



t
maxc
003.0
b
yt




bc
Maximum and Balanced Steel Ratios








max
max
max
5686.1
6375.0
8
1.5
008.0
0051.0
0021.0:420For




b
b
bb
yy MPaf

25-Feb-13
Tension steel strain check
Maximum steel ratio check
Neutral axis depth check
• Ensuring that steel ratio is less than or equal to the
maximum limit is equivalent to checking that tensile steel
strain is equal to or greater than 0.005
• This is also equivalent to checking that neutral axis depth
is limited such (using similar triangles):
19
dc
d
c
d
c
t
t
t
375.0005.0
375.0
8
3
005.0
003.0
003.0
max 






d
003.0
005.0t
c
• These three checks are equivalent.
• The tension steel strain is sufficient and more
informative about steel strain values.
• In case of many tension steel layers, the maximum
steel ratio check and the neutral axis depth check
may be misleading.
• It is therefore recommended to always use steel
strain check and discard the other two checks.
20
Tension steel strain check
Maximum steel ratio check
Neutral axis depth check
dct 375.0005.0 max  

25-Feb-13
Strength Reduction Factors 
Strength reduction factors () account for :
• Uncertainties in material strength
• Approximations in analysis and simplifications in equations
• Variations in dimensions and in placement of reinforcement
• Type of failure (Tension-control or compression-control)
Typical values of  :
• Tension controlled sections :  = 0.90
• Compression members and Tied Columns :  = 0.65
• Compression Spiral Columns :  = 0.70
• Flexure and compression (transition) :  = 0.65 (0.70) to 0.90
• Shear and torsion :  = 0.75
• Bearing on concrete :  = 0.65
21
22
Variation of  with Tensile Steel Strain
 The strength reduction factor varies in the transition region.
 Beams must have a tensile steel strain of at least 0.005 for
SBC code (0.004 for ACI code).

25-Feb-13
23
Types of Beams
Analysis of RC Beams
Determination of moment capacity
24
• Analysis of existing RC beams is somewhat different from
the analysis and check involved in the design stage.
• For new beams to be designed, all code provisions must be
enforced. If any condition is not satisfied, the design must
be repeated.
• For existing beams, the analysis must be completed and any
deficiencies reported. If for instance the section is not tension
controlled or not satisfying minimum steel condition, its
nominal and design moments must still be determined using
appropriate strength reduction factors.

25-Feb-13
25
Values of strength reduction factor  for beams
• Tension control condition is required in all beam design
problems, with a strength reduction factor ( = 0.90)
• In analysis problems however, other cases may be met.
• The value of the strength reduction factor depends on the
tensile steel strain at the extreme (bottom) layer:
65.0:controlnCompressioIf
005.0
25.065.0:zoneTransition005.0If
90.0:controlTension005.0If










yt
y
yt
ty
t
26
d
b
sA
N.A.
c ca 1 abfC c
'
85.0
'
85.0 cf
ss fAT 
The main difficulty in analysis problems is that it is not known
whether steel has yielded or not. Existing beams may have been
designed as over reinforced with steel not yielding at failure.
Yielding case is much simpler as the steel stress is then known and
constant fy
It is usually first assumed that steel has yielded and if turns out that
it has not, then a different method is used.
Nominal Moment in RC Beams

25-Feb-13
27

















22
:momentNominal
:continueyiedling,OK:If0030ianglesSimilar tr
850
850
1
a
dfAM
a
dTM
ε
c
cd
.ε
a
c
bf.
fA
afAabf.TC
ysnn
ytt
'
c
ys
ys
'
c


Nominal Moment – Tension Steel Yielding
d
b
sA
(a) Section
(b) Strains
(c) Stresses / Forces
N.A.
003.0
yt  
ca 1
abfC c
'
85.0
'
85.0 cf
ys fAT 
c
28
0600600850
)(600850600850
:givesand850850
600)200000(
2
1
2
11
1







dAcAcbf.
cdAcbf.
c
cd
Acbf.
TCcbf.abf.C
c
cd
ATMPaE
ss
'
c
s
'
cs
'
c
'
c
'
c
ss



Nominal Moment – Tension Steel Not Yielding
c
cd
.EAT
c
cd
EAfAT
ss
t
tssss
yt






0030
003.0
:ianglesSimilar tr
:ThenIf



d
b
sA
N.A.
003.0
yt  
ca 1
abfC c
'
85.0
'
85.0 cf
tssss EAfAT 
c
(a) Section
(b) Strains
(c) Stresses / Forces

25-Feb-13
29
Nominal Moment – Tension Steel Not Yielding





























22
:momentNominal
:yieldednothassteelionthat tensConfirm
600and003.0,:Deduce
1
4
1
2
:issolutionPositive
solutions.ith twoequation wQuadratic
85.0
600
with0
0600600850
1
1
'
2
2
1
a
dfAM
a
dTM
c
cd
εEf
c
cd
ca
P
dP
c
bf
A
PPdPcc
dAcAcbf.
ssnn
yt
tsst
c
s
ss
'
c




Analysis of a rectangular RC section
Steps in determining design moment
30
6Goto



















2
:yieldingOKIf4/
003.0strainsteelTensile/3
and
85.0
yieldingsteelAssume/2
:Check
4.1
,
4
Max,
:valuesratiosteelminimumandactualCompute1/
t
1
min
'
min
a
dfAM
c
cd
a
c
bf
fA
a
ff
f
bd
A
ysny
t
'
c
ys
yy
cs





25-Feb-13
Steps in determining design moment – Cont.
31
 
n
t
yt
y
yt
ty
t
ssnn
tssytt
c
s
yt
M
ε
ε.
εε
ε
a
dfAM
a
dTM
εEf
c
cd
ca
bf
A
P
P
dP
c








ismomentDesign/7
SBC)byRejected,controllednNot tensio:005.0(If
0.65:controlnCompressioIf
0050
25.065.0:zoneionIn transit005.0If
0.90:controlTension005.0If
:factorreductionSrength6/
22
:momentNominal
:check003.0,
85.0
600
with,1
4
1
2
:yieldingnotSteel:If5/
1
1
'































Problem 1
32
85.0MPa30MPa420 1  '
cy ff
600
375
284
75
675
 
OK
00333.000333.0,00326.0Max
420
4.1
,
4204
30
Max
4.1
,
4
Max
0109.0
600375
0.2463
:ratioSteel
0.246382
4
4:areaSteel
min
min
'
min
22




































yy
c
s
s
ff
f
bd
A
mmA
Determine the nominal and design moments of the beam
section shown.

25-Feb-13
33
mm3.127
85.0
2.108
mm2.108
3753085.0
4200.2463
85.0
yiedingsteelAssume
1







a
c
a
bf
fA
a
ff
'
c
ys
ys
0.90
controltensionandyieldingSteel
005.00111.0
0111.0
3.127
3.127600
003.0
003.0












yt
t
t
c
cd
kN.m508.24
kN.m71.5649.0:momentDesign
kN.m71.564N.mm1071.564
2
2.108
6004200.2463
22
6























n
n
n
n
ysn
M
M
M
M
a
dfA
a
dTM


Solution 1
3.127
003.0
7.472
s
600
375
284
Problem 2
section shown. fc’ = 20 MPa and fy = 420 MPa
34
525
600
mm350
283
 
OK
00333.000333.0,00266.0Max
420
4.1
,
4204
20
Max
4.1
,
4
Max
01005.0
525350
25.1847
:ratioSteel
25.184782
4
3:areaSteel
min
min
'
min
22




































yy
c
s
s
ff
f
bd
A
mmA

25-Feb-13
35
mkNM
a
dfAM
MOK
c
cd
mm
a
c
n
t
ysn
nyt
t
.03.3217.35690.0
control)(Tension90.0005.000726.0
kN.m7.356N.mm107.356
2
39.130
52542025.1847
2
:forContinue.assumedasyieldedhassteel0021.000726.0
00726.0
4.153
4.153525
003.0003.04.153
85.0
39.130
:yieldedhassteelCheck that
6
1


























525
350
283
ca 1
abfC c
'
85.0
'
85.0 cf
ys fAT 
2
a
d 
Solution 2: Assume steel yielding
mm39.130
3502085.0
42025.1847
85.0
25.184728
4
3
'
22







a
bf
fA
aTC
mmA
fATff
c
ys
s
ysysys


Problem 3
section shown. Same as previous problem but with one more
steel bar. fc’ = 20 MPa and fy = 420 Mpa
36
 
OK
00333.000333.0,00266.0Max
420
4.1
,
4204
20
Max
4.1
,
4
Max
0134.0
525350
0.2463
:ratioSteel
0.246382
4
4:areaSteel
min
min
'
min
22




































yy
c
s
s
ff
f
bd
A
mmA
525
600
mm350
284

25-Feb-13
37mkNM
a
dfAM
MOK
c
cd
mm
a
c
n
t
ty
ysn
nyt
t
.06.39616.453874.0
874.0
0021.0005.0
0021.00047.0
25.065.0
005.0
25.065.0:Transition005.00047.0
kN.m53.164N.mm1016.453
2
86.173
5254202463
2
:forContinue.assumedasyieldedhassteel0021.00047.0
0047.0
54.204
54.204525
003.0003.054.204
85.0
86.173
y
y
6
1



































Solution 3: Assume steel yielding
:strainsteelCheck
mm86.173
3502085.0
4200.2463
85.0 '






a
bf
fA
aTC
fATff
c
ys
ysysyt 
525
350
284
ca 1
abfC c
'
85.0
'
85.0 cf
ys fAT 
2
a
d 
Problem 4
• The reinforcement of 2945 mm2 is from
six 25-mm bars which are arranged in
two layers but are assembled (lumped)
together in a single compact layer.
38
Determine the nominal and design moments
of the beam section shown.
fc’ = 20 MPa and fy = 420 MPa
510
250
2945
90
600
2
mm
  OK00333.000333.0,00266.0Max
420
4.1
,
4204
20
Max
4.1
,
4
Max
0231.0
510250
2945
:ratioSteel
minmin
'
min


























yy
c
s
ff
f
bd
A

25-Feb-13
39
mm4.342
85.0
04.291
mm04.291
2502085.0
4200.2945
85.0
yiedingsteelAssume
1







a
c
bf
fA
a
ff
'
c
ys
ys
mmcammc
cP
bf
A
P
P
dP
c
c
cd
c
s
yt
t
82.26455.31185.055.311
1
135.489
5104
1
2
135.489
135.489
85.02502085.0
2945600
85.0
600
with,1
4
1
2
yieldingNot0147.0
00147.0
4.342
4.342510
003.0003.0
1
1
'


































Solution 4
510
250
2945
90
600
2
mm
40
kN.m.24762kN.m98.42465.0:momentDesign
0.65:controlnCompressio
kN.m98.424N.mm1098.424
2
82.264
51018.3820.2945
22
6






















n
yt
n
ssn
M
M
a
dfA
a
dTM


Solution 4 - Continued
)(18.3820019109.0200000
yieldingNotOK0019109.0
55.311
55.311510
003.0003.0
82.26455.311
ytss
yt
fMPaEf
c
cd
mmammc









This beam is over reinforced (compression controlled) and is not
allowed by SBC / ACI codes. The two steel layers were lumped at
their centroid but lumping is allowed only if all layers are yielding.
Non yielding with many layers can only be analyzed using the
strain compatibility method described later.

25-Feb-13
41
Lumping of tension steel layers at centroid
• If tension steel is arranged in two or more layers, these can be,
in the calculations, replaced by a single steel area at their
centroid. This Layer lumping (assembly) simplifies analysis and
design equations. Standard beam design cannot in fact deal
with more than one tension steel layer.
• As layer strains are different, lumping is only justified if all
tension layers have yielded (and have the same yield stress).
• Tension steel depth d must therefore be computed at the
centroid of the layers.
• Yielding must be checked at the least tensioned layer with
minimum depth dmin
• It is therefore unsafe to check yielding at the centroid
• Performing tension control check at the centroid is
uneconomical. It should be performed at the most tensioned
extreme bottom layer.
42
0.003
d
dt
c
Depths Strains
005.0t
yε min
dmin
s
Lumping of tension steel layers at centroid






005.0
:layersMany
005.0:layerOne
min
t
y
t
ε



Required steel
strain checks :
ts
ts
ε
ε




min
min
:layersMany
:layerOne
centroidsteelAt tension:,
layer(bottom)max.depthAt:,
layerdepthminimumAt:,
:usedNotations
minmin
s
tt
εd
εd
εd

25-Feb-13
Determine design moment for the shown section with six
20-mm bars in two layers. Bar / layer spacing is 30 mm.
43
MPa420andMPa20  y
'
c ff
250
206
650
Problem 5
mm
dd
d
mmdd
dSddd
mmdd
d
dhdd
bl
t
b
st
565
2
:centroidatdepthsteelEffective
5402030590
:layerSecond
590
2
20
1004650
2
cover:layerbottomFirst
:depthsSteel
21
min2
1min2
1
1







Solution 5
44
 
OK
00333.000333.0,00266.0Max
420
4.1
,
4204
20
Max
4.1
,
4
Max
0133.0
565250
96.1884
:ratioStee
96.188402
4
6:areaSteel
min
min
'
min
22




































yy
c
s
s
ff
f
bd
A
mmA
250
206
565
mm15.219
85.0
28.186
mm28.186
2502085.0
42096.1884
85.0
depthblockStress
:steelofyieldingAssume
1
1 






a
cca
bf
fA
a '
c
ys


25-Feb-13
45
layer.steelbottomattestcontroltensionperformBetter
controlled-nnot tensioissectionthecentroid,At the
005.00047.0
15.219
15.219565
003.0003.0
:centroidsteelatcheckStrain:Important
0.90controlTensionOKandyieldOK
005.000507.0
15.219
15.219590
003.0003.0
:layer)(bottomdepthmaximumatcheckcontrolTension
0021.00044.0
15.219
15.219540
003.0003.0
:depthminimumatstrainsteelofcheckYield
min
min


















c
cd
c
cd
OK
c
cd
s
y
t
t
y




Solution 5 Continued
46
kN.m20.363
kN.m56.3739.0:momentDesign
kN.m56.373N.mm1056.373
2
28.186
56542096.1884
22
6






















n
n
n
ysn
M
M
M
a
dfA
a
dTM


15.219
003.0
0044.0
250
206
565
00507.0
0047.0
This example shows the importance
(economy) of performing the tension
control check at the extreme bottom
layer of tension steel.

25-Feb-13
Determine the design moment for the shown section
47
MPa420andMPa27  y
'
c ff
Problem 6
mm
dd
d
mmdd
mmdd t
608
2
580:layerSecond
636:layerbottomFirst
21
min2
1





  OK00333.000333.0,00309.0Max
420
4.1
,
4204
27
Max
4.1
,
4
Max
0270.0
608300
0.4926
0.492682
4
8
minmin
'
min
22

































yy
c
s
s
ff
f
bd
A
mmA
300
288
700
Solution 6 (assume steel yielding)
48
ys
ty
t
t
y
'
c
ys
c
cd
c
cd
c
cd
a
c
bf
fA
a
























00216.0
53.353
53.353608
003.0003.0:straincentroidSteel:Important
ionIn transit005.000240.0
53.353
53.353636
003.0003.0
justifiednotLumpingyieldingNo00192.0
53.353
53.353580
003.0
003.0:depthminimumatstrainsteelofcheckYield
mm53.353mm50.300
3002785.0
4200.4926
85.0
depthBlock
min
min
min
1
Steel is not yielding at the minimum depth layer although it is
yielding at the centroid. Lumping is therefore unjustified and would
unsafely deliver a higher moment capacity. This example proves
that it is unsafe to perform the yield check at the centroid.

25-Feb-13
49
mkNM
a
dfAM
n
ysn
.2.640kN.m05.947676.0:momentDesign
676.0
0021.0005.0
0021.00024.0
25.065.0transitionIn005.000240.0
kN.m05.947N.mm1005.947
2
5.300
6084200.4926
2
tt
6




















This example shows the importance
(safety) of performing the steel yield
check at the least tensioned steel
layer (with minimum depth).
The unsafe (over estimated) moment
capacity would be as :
This result is of course unsafely false because of an unjustified use
of layer lumping. The exact nominal moment obtained using the
strain compatibility method is 935.76 kN.m
53.353
003.0
00192.0
300
288
608
00240.0
00216.0
Determine design moment for the shown section
(shallow beam very popular in KSA) with nine 20-mm
bars in three layers. Layer spacing is equal to 30 mm.
It is obvious this is not a correct bar arrangement as the
beam width can accommodate more than three bars.
However design and / or execution errors are
unfortunately always possible.
50
 
0026.0and
802.065.0,008.009.1Max
MPa520andMPa36
y
1




 '
c
y
'
c
f
ff
Problem 7 – Special and Important
500
209
380

25-Feb-13
51
Solution of Problem 7
mmd
ddd
d
mmdd
dSddd
mmd
dSdd
mmdd
d
dhdd
bl
bl
t
b
st
270
3
2202030270
:layerThird
2702030320
:layerSecond
320
2
20
1004380
2
cover:layerbottomFirst
:depthsSteel
2
321
min3
2min3
2
12
1
1









500
209
380
Solution 7
52
 
OK
00284.000269.0,00284.0Max
520
4.1
,
5204
35
Max
4.1
,
4
Max
0209.0
270500
4.2827
:ratioStee
4.282702
4
9:areaSteel
min
min
'
min
22




































yy
c
s
s
ff
f
bd
A
mmA
mm81.119
802.0
09.96
mm09.96
5003685.0
5204.2827
85.0
depthblockStress:yieldingsteelAssume
1





a
ca
bf
fA
a '
c
ys
500
209
380

25-Feb-13
53
ys
t
t
t
y
c
cd
c
cd
c
cd






















00376.0
81.119
81.119270
003.0003.0
:centroidsteelatcheckStrain:Important
ControlTensionOK005.000501.0
00501.0
81.119
81.119320
003.0003.0
OKNOTCheckYield0026.000251.0
00251.0
81.119
81.119220
003.0003.0
:depthminimumatstrainsteelofcheckYield
min
min
min
54
The section is tension-controlled (accepted by SBC) as the bottom
strain is just greater than the 0.005 limit.
However, although steel has yielded at the centroid, the strain in
the upper layer is less than the yield limit.
The previous calculations, based on the assumption of yielding of
all steel, are thus incorrect and unsafe.
It is therefore more exact and safer to perform the yield check at
the upper layer with minimum depth.
ys
t
y








00376.0
ControlTensionOK
005.000501.0
OKNOTYield
0026.000251.0min
8.119
003.0
00251.0
500
209
270
00501.0
00376.0

25-Feb-13
55
kN.m70.293kN.m33.3269.0:momentDesign
kN.m33.326N.mm1033.326
2
09.96
2705204.2827
22
6





















n
n
ysn
M
M
a
dfA
a
dTM

This example shows the
importance (safety) of
performing the tension
control check at the upper
layer of tension steel.
This incorrect solution for
the nominal and design
moments is :
8.119
003.0
00251.0
500
209
270
00501.0
00376.0
The exact solution based on the actual strain in every layer, would
deliver lower values of moments. As the strain in the upper layer is
close to yield, the exact solution should be just slightly lower.
56
Strain Compatibility Solution 7
• The strain compatibility method is a very powerful analysis tool
for complex situations, with many steel layers.
• It is based on the linear variation of strains resulting from the
assumption of sections remaining plane after bending.
• The stress at any level or layer is related to the strain through the
material model (stress-strain curve). No layer lumping is used.
• All strains are expressed in terms of the neutral axis depth c
which is the main problem unknown.
c
003.0
3s
500b
2033 
1s
2s
c
cd
i
i
si

 003.0
:layerAt


25-Feb-13
57
Strain Compatibility Solution 7 – Cont.
• The method is more effective if a good initial assumption is
made about yielding (or not) of the various steel layers.
• It is reasonable to assume that at failure the bottom two layers
have yielded and that the third is in the elastic range (stress
proportional to strain through Young’s modulus). The strains and
forces are therefore as shown in the figure.
c
003.0
ys  3
500
2033 
ys  1
ys  2
ca 1
'
85.0 cf
33 sss EA 
ys fA1
ys fA 2
abfc
'
85.0
Section Strains Forcesc
cdi
si

 003.0
58
• The two unknown forces are the concrete compression and the
tension in the third layer. They are both expressed in terms of
the neutral axis depth (using 200000 MPa for steel modulus) :
c
cd
A
c
cd
EAEATcbfabfC sssssssccc



 3
3
3
33331
''
600003.085.085.0 
c
003.0
ys  3
500
2033 
ys  1
ys  2
ca 1
'
85.0 cf
33 sss EA 
ys fA1
ys fA 2
abfc
'
85.0
Section Strains Forcesc
cdi
si

 003.0

25-Feb-13
59
• We now express the force equilibrium :
c
cd
AEATcbfabfC sssssccc

 3
33331
''
60085.085.0 
c
cd
AfAfAcbfTTTC sysyscsssc

 3
3211
'
321 60085.0 
• Remark: If more than one layer has not yielded, its strain, stress
and force are again expressed in terms of the neutral axis depth
(c) and the latter will remain the only unknown in the final
equilibrium equation.
• We multiply by c and after assembling, we obtain a quadratic
equation :
   
  060060085.0
60085.0
33321
2
1
'
3321
2
1
'


dAcAfAfAcbf
cdAcfAfAcbf
ssysysc
sysysc


60
• Substitution with the given data leads to :
002546.000381.000507.0003.0
0.119:issolutionPositive
2.20463.41696
0632.10138795.33
012440709632.4146906.12270
321
2
2







sss
i
si
c
cd
mmc
cc
cc

 
mmdmmdmmd
mmAAA
mmbMPafMPaf
dAcAfAfAcbf
sss
yc
ssysysc
220270320
478.942
802.050052036
060060085.0
321
2
331
1
'
33321
2
1
'







25-Feb-13
61
• The assumed steel strain conditions are confirmed. The third
layer has not reached the yield limit but is very close to it.
• If the assumed yielding (or not) condition is violated in any
layer, then new calculations must be carried out.
• The strain in the bottom layer confirms that the section is
tension controlled :
• Substitution gives the forces, which must satisfy equilibrium.
002546.000381.000507.00.119 321  sssmmc 
kNTkNTTkNC
c
cd
ATfATfATcbfC
sssc
ssyssysscc
95.47909.49020.1460
60085.0
321
3
3322111
'


 
• Equilibrium is checked. The sum of the steel tensile forces is
equal to concrete compression force.
62
• The nominal moment is
better determined as the
moment of the steel
tensile forces about the
compression centroid.
• Using kN and meter units :
'
85.0 cf
33 sss EA 
ys fA1
ys fA 2
abfc
'
85.0
44.95
1ca 
'
85.0 cf
95.479
09.490
09.490
20.1460
mkNMmkNM
M
a
dT
a
dT
a
dTM
nn
n
sssn
.55.29206.32590.0.06.325
2
09544.0
22.095.479
2
09544.0
27.009.490
2
09544.0
32.009.490
222
332211








































• This exact solution is less than but close to the previous false one
because the third layer was very close to yielding. Despite non
yielding of the top layer, the section is still tension controlled.

25-Feb-13
6363
Lumping of many tension steel layers
Caution
• Lumping all steel layers at the centroid
may only be valid for normal beams not
exceeding three to four layers.
• For deep beams or columns and walls
with many steel layers, and for beams
with mid-height side (skin) bars, lumping
cannot be used and appropriate methods
must be used (strain compatibility).
• The two shown skin bars (imposed by
codes of practice for deep beams) must
not be included in the lumping process.
63
Skin bars
RCcolumnRCdeepbeam
Homework Problems
64
1/ Determine the design moment capacity for the beams shown.
f’c = 30 MPa , fy = 420 MPa.
600
375
363
75
675
600
375
283
75
675
Beam 1 Beam 2
2/ Re-solve problem 6. Obtain the given exact solution using the
strain compatibility method.

Lec04 Analysis of Rectangular RC Beams (Reinforced Concrete Design I & Prof. Abdelhamid Charif)

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