Flexural Analysis & Design of Rectangular Beams:
EC2 & ACI318-19 Compared
Kamal J. Bsaibes
Structural Engineer
The increasing complexity of a large number of projects and the development of new
techniques have led to a rapid and significant extension of the regulations on the basis of the
experience acquired and the results of the numerous researches carried out in several
countries. Today, the engineer is therefore obliged to comply with regulations which are not
the same everywhere and given the large share of activity of design offices abroad, engineers
must also have a good knowledge of the main foreign regulations. Only the big nations have
developed their own regulations and almost all the other countries have chosen one of the
two major reference text in reinforced concrete structures : the ACI 318-19 and the
Eurocode2 (EC2). We start by comparing “Simple bending in reinforced rectangular beam”
with numerical comparison for each subject. These comparisons leads to general conclusions
about the cases treated. The following text accompanied by an application
https://www.microsoft.com/store/productId/9NFXL4VWSDX0, is added to other texts in the
literature in order to acquire more profit on the subject.
EUROCODE 2
ACI318-19
Hypotheses
The following limitations and simplifications are taken into account in this comparative
study:
• Service loads are equal;
• The value of the coefficient cc  1 ( cc  0.85 some national annexes) ;
We will consider in the following : cc  1 ;
• For concrete : fck ( EC2)  fc' ( ACI ) and for steel : f yk ( EC 2)  f y ( ACI ) ;
• Plane sections before bending remain plane after bending;
• Unit strain in the concrete is proportional to the distance from the neutral axis;
• The tensile strength of the concrete is ignored;
• The adhesion between steel and concrete must be completely perfect. i.e. there is no relative
displacement between concrete and steel;
• The stress in steel varies linearly according to the unit strain up to the elastic limit; it
remains constant after this point while the unit strain increases (EC2 allows the use of a
stress-strain diagram of horizontal top branche) ;
• The ultimate concrete unit strain must be limited to 0.003 (ACI 318-19) and to 0.0035
 cu3  according to EC2.
EUROCODE 2
ACI318-19
Stress distribution
When the section is not fully compressed, the
The stress distribution in compressed
"parabola-rectangle" diagram can be replaced
concrete can be considered as rectangular,
by the simplified rectangular diagram defined
trapezoidal, parabolic, or any other form
below:
giving results in agreement with the test
results. Rectangular Diagram:
fcd  fck  c
fck  50MPa :

1  0.85
  0.80 ;   1

50MPa
fck  90MPa :

λ  0.8   fck  50  / 400
η  1   fck  50 / 200
h = overall depth of the section
17 MPa  fc'  28MPa :
28MPa
fc'
 1  0.85 

7
fc'  55 MPa : 1  0.65
h = overall depth of the section
b = breadth of the section
d = effective depth,
x = depth to the neutral axis
b = breadth of the section
As = area of tension reinforcement
c = depth to the neutral axis
 s = strain in steel

0.05 fc'  28
d = effective depth,
 cu 3 = maximum strain in the concrete
55 MPa :
As = area of tension reinforcement
 c = maximum strain in the concrete
t = strain in steel

EUROCODE 2
ACI318-19
Strain Compatibility
With same ordinate, the strain of steel is
With same ordinate, the strain of steel is
equal to the strain of concrete:
equal to the strain of concrete:
s
d x
 d x 

  s   cu3 

 cu3
x
 x 
t d  c
 d c 

 t   c 

c
c
 c 
Let :

 cu3
x

d  cu3   s
 1 
 et
  
 s   cu3 
Let :

c
c


d  c  t
 1 
 et
  
  

 1 
t   c 
c  s 
  

 1 
 c  t 
- For fck  50MPa ;  cu3  0.0035 ;
For εc  0.003 ; ε t  f y Es ;
E s  200000 MPa
E s  200000 MPa
 
x
0.0035
700


d 0.0035  f yd E s 700  f yd
For f yk  500MPa ;
f yd  500 1.15  435 MPa

x
700

 0.617
d 700  435
Hence, to ensure yielding of the tension steel
at the ultimate limit state : x  0.617d
for fck  50MPa and f yk  500MPa
At the ultimate limit state it is important that
 
For f y  500MPa ;

neutral axis to :

x
 0.45 .
d
c
600

 0.546
d 600  500
Hence, to ensure yielding of the tension steel
at the ultimate limit state : c  0.546d
for f yk  500MPa
To ensure ductility, ACI limits the depth of
neutral axis to :
member sections in flexure should be ductile.
For this purpose, EC2 limits the depth of
c
0.003
600


d 0.003  f y Es 600  f y

c

d
0.003
.
 fy

0.003  
 0.003 
 Es

EUROCODE 2
ACI318-19
Design of a singly reinforced rectangular section
C   x.cc. fcd .b
C  0.85 fc' b1c  0.85 fc' 1bd
with : x   d
with : c   d
The tension force of steel is: T  As fs
The tension force of steel is: T  As fs
Taking :
z d
a
 

 z  d1
2
2 

The balance of normal efforts result in :
Taking :
zd
The balance of normal efforts result in :
T=C
T=C
As fs   x.cc. fcd .b
x 
As . f s

As . f s
cc . fcd .b . fcd .b
.( x d ) 
As . f s
 . fs

b.d. . fcd  . fcd
a
 

 z  d1 1 
2
2 

As f s  0.85 fc' b   1c 
(  cc  1)
( 
Steel percentage :
. x d .. fcd . .. fcd
A
 s 

b.d
fs
fs
As
)
b.d
 1c 
 1 .(c d ) 
As . f y
0.85. fc' .b
As . f s
b.d.0.85.
f c'

 . fs
0.85.
f c'
( 
As
)
b.d
Steel percentage :

 1 .  c d  .0.85. fc'  1 . .0.85. fc'
As


b.d
fs
fs
EUROCODE 2
ACI318-19
Design of a singly reinforced rectangular section
Resisting internal moment  C  z  T  z
Resisting internal moment:
Mu  M Ed  C  z   xcc fcd bz
Mu ø  M n  C  z  T  z
The lever arm z is such that:
Mu ø  Mn  C  z  1c(0.85 fc' )bz
  
z  d  0.5  x  d  0.5  d  d  1 
2 

The lever arm z is such that:
 

M u   d cc fcd b  d 
d
2 

Mu  ø( 0.85 fc' )bd 21  1  1 2  
  
M u   cc bd 2 f cd  1 

2 

u 
Mu
bd 2 fcd
 
z  d  0.5  1c  d  0.5  1 d  d  1  1 
2 

Mu
ø ( 0.85
 

  cc  1 
2 

u 
For fck  50MPa :   0.8 ; η=1 ;  cc =1
u 
Mu
bd 2 fcd

Mu  M Ed

et z  d  1  0.4 
Si  s  f yd E s  fs   s Es

Si  s  f yd E s  f s  f yd 
2
Mu
ø( 0.85
fc'
1
1


  1  1  1 
2 
)bd

2
1 
a   1 c   1 d
et
1  2 u

z d
a


 d1 1 
2
2 

Mu ø  T  z  As fs z  As 
M
 T.z  As s z  As  Ed
sz



  1  1  1 
2 
)bd


 0.8  1  0.4  
  1.25 1  1  2 u
fc'
f yk
s
Mu
øzf s

Si  s  f y E s  fs   s Es

Si  s  f y E s  f s  f y
EUROCODE 2
ACI318-19
Determination of As,min et de As,max
A minimal percentage of steel is required for bending elements by the two codes, for
controlling the opening of cracks. Both codes impose a maximum percentage of steel to
ensure adequate ductility.
Minimal percentage:
Minimal percentage:


f
As,min  max 0.26 ctm bd ; 0.0013bd 
f yk



f'
1.4 
As,min  Max  0.25 c bd ;
bd


fy
fy


min  As,min bd
This value of As ,min does not apply, if
As,mis en place  1.33 As,calculé .
min  As,min bd
In order to guarantee sufficient ductility of the structure elements at ultimate state (plastic
behavior), the EC2 and ACI define the following limits:
For fck  50MPa ;
s  (
1    
 s  c 
  0.0043
  
 c  3.5
fy
Es
 0.003) to avoid excessive steel
section :
  x d  0.45
 c  0.003   s 
max  0.45
  c d 
For 50  fck  90 MPa ;
max 

0.46
1.25 *  0.6  0.0014 /  cu 3 

max 
fy
3 1
 (  0.003)
1000 
Es
0.003
 fy

0.003  
 0.003 
 Es

0.003
fy
Es
 0.006
These limitations of neutral axis can be converted into limiting moments that sections can
support without addition of compressed steel.
EUROCODE 2
ACI318-19
Determination of As,max
Percentage limit (not to have
Percentage limit (not to have
compression reinforcement) :
compression reinforcement) :
l 
 .max . . fcd
l 
fs
Maximum tension and compression steel
area should not exceed 4% of the gross
1 .max .0.85. fc'
l  0.85  1
section area.
f ck (MPa)
fctm (MPa)
min (EC2)%
ou :
fs
min (ACI)%
fc'
fy
c
c  (
fy
Es
 0.003)
 l (EC2)%
 l (ACI)%
f yk  500 (MPa)
25
2.6
0.14
0.28
1.38
1.28
30
2.9
0.15
0.28
1.66
1.50
35
3.2
0.17
0.30
1.93
1.68
40
3.5
0.18
0.32
2.21
1.83
45
3.8
0.20
0.34
2.49
1.97
50
4.1
0.21
0.35
2.76
2.08
55
4.2
0.22
0.37
3.45
2.15
60
4.4
0.23
0.39
3.61
2.34
70
4.6
0.24
0.42
3.86
2.73
80
4.8
0.25
0.45
4.03
3.12
90
5
0.26
0.47
4.12
3.51
The previous table shows that the minimum percentage of steel required by the ACI code is
greater than that of the EC2; while the maximum percentage of steel required by EC2 code, to
ensure ductility, is greater than that of ACI .
EUROCODE 2
ACI318-19
Moment limit values
We compare the moments limit of the two regulations for :
fck ( EC2)  fc' ( ACI )  25 MPa and f yk ( EC 2)  f y ( ACI )  500MPa
Let: G = dead load and Q = live load; total load : P = G + Q.
By taking into account the load factors for each regulation, the moments limit values are
calculated according to different values of Q P ; the value of the moment calculated, is
related to the total load.
fck  25 MPa  fcd  25 1.5  16.67 MPa
max  0.8max  1  0.4max  ;
max  0.8  0.45  1  0.4  0.45   0.295
Ml EC 2
fcd bd 2  max  0.295 
Ml EC 2 16.67bd 2  0.295 
max 
0.003
 0.353
f y E s  0.006


max  max  1  1 
 max  1 
2
  0.247

Ml ACI  0.9  0.85  bd 2 fc'  max  0.247
Ml EC 2 19.13bd 2  0.247 
Ml EC 2  4.92bd 2 ;
1.35G  1.5Q  1.35  P  Q   1.5Q
0.15Q 

1.35G  1.5Q  1.35 P  1 

1.35 P 

Ml ACI   4.72bd 2 .
1.2G  1.6Q  1.2  P  Q   1.6Q
0.4Q 

1.2G  1.6Q  1.2P  1 
1.2P 

Q P
M l EC 2 
M l ACI 
Ml ACI  Ml EC 2
0
3.644
3.933
1.080
0.20
3.565
3.688
1.035
0.25
3.546
3.631
1.025
0.30
3.527
3.578
1.015
0.35
3.508
3.522
1.006
0.40
3.489
3.471
0,996
0.45
3.471
3.420
0.987
0.50
3.453
3.371
0.978
In practice the ratio Q P varies between 0.2 and 0.5, and it is noted that the sections in
bending at EC2 require addition of compression steel for moments limit lower than the ACI
moments limit.
EUROCODE 2
ACI318-19
Applications
Design the following beam for : MG  90kNm ; MQ  30kNm .
fck ( EC2)  fc' ( ACI )  25 MPa
et f yk ( EC 2)  f y ( ACI )  500MPa .
Ultimate moment:
Ultimate moment:
Mu  1.35  90  1.5  30  166.5kNm
Mu  1.2  90  1.6  30  156kNm
As : fck  50MPa donc   0.8 et η=1 ;
fc'  25MPa  β1  0.85
u 
Mu
bd 2 fcd


166.5  10 3
0.30  0.50 2  16.67

  1.25 1  1  2 u  0.179
 0.133
max  0.45
μu 
Mu
ø(0.85fc' )bd2
μu 
124  103
 0.108
0.9  0.85  25  0.30  0.502
So no need for compression steel.
 1 
 1  0.179 
  0.0035 
  0.016
 0.179 
  
s  c 
ξ


1
1  1  2μ u  0.135
0.85
 s  0.016
 y  500 ( 1.15  2  10 5 )
ξ max  0.003 (f y Es  0.006)  0.353
 s  0.016
 y  0.00217
ξ  0.135
  s  f yd  500 1.15  435 MPa
ξ max  0.353
So no need for compression steel.
 1 
 1  0.135 
z  d  1  0.4   0.50 1  0.4  0.179   0.464m  s   c     0.003  0.135   0.019




As 
Mu 166.5  103

 0.00082m 2
 s z 435  0.464
As  8.2cm2
 s  0.019  y  500 2  10 5  0.0025
  s  f y  500MPa ;
z  d  1  0.5  1 
z  0.50 1  0.5  0.85  0.135   0.471m
As 
Mu
156  103

 0.00074m2
øzf y 0.9  0.471  500
A s  7.4cm2
https://www.microsoft.com/store/productId/9NFXL4VWSDX0
EUROCODE 2
ACI318-19
For the same beam above; we consider : MG  90kNm ; MQ  60kNm
Ultimate moment:
Ultimate moment:
Mu  1.35  90  1.5  60  211.50kNm
Mu  1.2  90  1.6  60  204kNm
As : fck  50MPa donc   0.8 et   1
fc'  25MPa  β1  0.85
u 
Mu
2
bd fcd

211.5  10 3
0.30  0.50 2  16.67


  1.25 1  1  2 u  0.234
 0.170
max  0.45
μu 
Mu
ø(0.85fc' )bd2
μu 
204  103
 0.142
0.9  0.85  25  0.30  0.502
So no need for compression steel.
 1 
 1  0.234 
  0.0035 
  0.011

 0.234 




1
1  1  2μ u  0.181
0.85
s  c 
ξ
 s  0.011  y  500 ( 1.15  2  10 5 )
ξ max  0.003 (f y Es  0.006)  0.353
 s  0.011  y  0.00217
ξ  0.181
  s  f yd  500 1.15  435 MPa
z  d  1  0.4   0.50  1  0.4  0.234   0.453m
As 
3
Mu 211.5  10

 0.001073m2
 s z 435  0.453
As  10.73cm2
ξ max  0.353
So no need for compression steel.
 1 
 1  0.181 
  0.003 
  0.014

 0.181 


s  c 
 s  0.014  y  500 2  10 5  0.0025
  s  f y  500MPa
z  d  1  0.5  1 
z  0.50 1  0.5  0.85  0.181  0.462m
As 
Mu
204  103

 0.00098m2
øzf y 0.9  0.462  500
A s  9.8cm2
The ACI gives steel areas lower than the EC2 by about 10%. This result is still general,
especially for weakly reinforced beams,
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EUROCODE 2
ACI318-19
For the same beam above; we consider : MG  170kNm ; MQ  90kNm
Ultimate moment:
Ultimate moment:
Mu  1.35  170  1.5  90  364.5kNm
Mu  1.2  170  1.6  90  348kNm
Comme : fck  50MPa donc   0.8 et η=1
fc'  25MPa  β1  0.85
u 
Mu
2
bd fcd


364.5  103
2
0.30  0.50  16.67

  1.25 1  1  2 u  0.444
 0.292
max  0.45
μu 
Mu
ø(0.85fc' )bd2
μu 
348  103
 0.243
0.9  0.85  25  0.30  0.502
So no need for compression steel.
 1
s  c 
 

 1  0.444 
  0.0035 
  0.0044
 0.444 

 s  0.0044
 y  500 ( 1.15  2  10 5 )
 s  0.0044  y  0.00217
  s  f yd  500 1.15  435 MPa
ξ


1
1  1  2μ u  0.333
0.85
ξ max  0.003 (f y Es  0.006)  0.353
ξ  0.333
ξ max  0.353
So no need for compression steel.
 1 
 1  0.333 
z  d  1  0.4   0.50  1  0.4  0.444   0.411m  s   c     0.003  0.333   0.006




As 
Mu 364.5  103

 0.00204m 2
 s z 435  0.411
As  20.4cm2
 s  0.006  y  500 2  10 5  0.0025
 s  f y  500MPa
z  d  1  0.5  1 
z  0.50 1  0.5  0.85  0.333  0.429m
As 
Mu
348  103

 0.0018m2
øzf y 0.9  0.429  500
A s  18cm2
For heavily reinforced beams, the ACI also gives smaller steel areas than the EC2. It follows
from the previous calculations that the EC2 always leads to steel areas larger than the ACI, this
difference being able to reach 15% for values of moments close to moments limit.
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EUROCODE 2
ACI318-19
Remark (1)
Let: G = dead load and Q = live load; total load : P = G + Q.
If we consider as an indication that, for the same value of the load P  G  Q :
- Steel according to EC2 are almost proportional to the quantity:
0.15Q 

1.35G  1.5Q  1.35 P  1 

1.35 P 

a 
- Steel according to ACI are almost proportional to the quantity:
0.4Q 

1.2G  1.6Q  1.2P  1 
1.2P 

b
Q P
0
0.1
0.2
0.3
0.4
0.5
0.6
 a b
1.125
1.101
1.078
1.057
1.037
1.018
1.00
It is noted that, for current values of the ratio Q P , varying between 0.2 and 0.5, the difference
between the steel areas given by the EC2 and the ACI decreases as Q P increases.
EUROCODE 2
ACI318-19
Remark (2)
We consider a beam of length l, uniformly loaded, of width b = 30cm and effective height 50cm.
MG et M Q being respectively the bending moment due to the dead loads and the bending moment
due to the live loads. The steel areas is calculated according to EC2 and ACI for different values of.
Q P with P  G  Q ; then we deduct the difference in percentage.
fck ( EC2)  fc' ( ACI )  25 MPa et f yk ( EC 2)  f y ( ACI )  500MPa
MG
MQ
Mu (EC2)
Mu (ACI)
As (EC2)
As (ACI)
Différence
(kNm)
(kNm)
(kNm)
(kNm)
(cm2)
(cm2)
(%)
0
80
0.00
108
96
5.208
4.42
17.83
0.1
80
8.89
121
110
6.657
5.87
13.41
0.2
80
20.00
138
128
8.544
7.75
10.25
0.3
80
34.29
159
151
11.04
10.29
7.29
0.4
80
53.33
188
181
14.67
13.80
6.30
0.5
80
80.00
228
224
20.26
19.29
5.03
Q P
We also notice that for the current values of the ratio Q P , the difference between the results given
by the EC2 and the ACI decreases when Q P increases.
EUROCODE 2
ACI318-19
Resistant moment of a section with known tension reinforcement
The balance of the compression force in
By writing the balance of the compression force in
concrete and the tensile force of steel gives:
concrete and the tensile force of steel we will have
C T
the nominal resisting moment which is equal to:
C   x.cc. fcd .b ; cc  1
a

Mn  A s f y  d  
2


The tensile force of steel is: T  f yd .As
Pour fck  50MPa :   0.8 et η=1 ;
0.8x. fcd .b  f yd .As  x 
f yd .As
0.8. f cd .b
So the resisting moment of the section is:
M sr
0.8x 

 T  z  f yd .As  d 
2 

M sr
As f yd 

 f yd .As  d 
 ;   As bd
2
fcd .b 

As  bd ; M sr
bdf yd 

 f yd .bd  d 

2 fcd .b 

 f yd

M sr   f yd bd 2  1 
2 fcd




 f yk 

 0.87  f yk  1 

2
1.533 fck 
bd

M sr
Avec :
a
A s .f y
0.85.fc' .b
Then we calculate øMn avec ø=0.9
So the resisting moment of the section is:
a

øM n  øAs f y  d  
2


As . f y 
øMn  øAs f y  d 
 ;   As bd

1.7 fc' .b 


bdf y 
As  bd ; øMn  øf y bd  d 


1.7 fc' .b 


 fy
øMn  ø  f y bd 2  1 
 1.7 f '
c

øMn
bd
2

 ; Mu  øMn


 fy
 ø fy  1 
 1.7 f '
c




In this calculation, it has been assumed that the steel has reached its elastic limit before the concrete
reaches its ultimate strain of 0.0035 or  cu 3 for (EC2) and 0.003 for (ACI). This hypothesis must be
verified by calculating the steel strain ε s for ultimate concrete strain.
EUROCODE 2
ACI318-19
Comparison of resisting moments
Comparison of resisting moments Mrésistant bd 2 according to the steel percentage  , for
f yk ( EC 2)  f y ( ACI )  500MPa and for different values of fck ( EC2)  fc' ( ACI ) .



Values of Mrésistant bd 2 MN m 2 for rectangular diagram
25MPA
30MPa
35 MPa
40MPa
50MPa
EC2
ACI
EC2
ACI
EC2
ACI
EC2
ACI
EC2
ACI
0.003
1.254
1.302
1.262
1.310
1.269
1.316
1.273
1.320
1.279
1.362
0.005
2.033
2.118
2.057
2.140
2.074
2.156
2.086
2.167
2.104
2.184
0.007
2.767
2.891
2.813
2.934
2.846
2.965
2.871
2.988
2.906
3.02
0.009
3.455
3.621
3.532
3.693
3.587
3.744
3.628
3.782
3.685
3.836
0.011
4.098
4.309
4.213
4.416
4.295
4.492
4.356
4.550
4.442
4.630
0.013
4.696
4.880
4.856
5.104
4.970
5.211
5.056
5.291
5.176
5.403
Note that the values given by ACI are higher than those of EC2. It also shows that EC2 is more
conservative than ACI concerning the bending resisting moment .
EUROCODE 2
ACI318-19
Neutral
Comparison of the values of   x d   c d  according to the steel percentage  , for
f yk ( EC 2)  f y ( ACI )  500MPa and for different values of fck ( EC2)  fc' ( ACI ) .

Valeurs de   x d   c d  pour un diagramme rectangulaire
25MPA
30MPa
35 MPa
40MPa
50MPa
EC2
ACI
EC2
ACI
EC2
ACI
EC2
ACI
EC2
ACI
0.003
0.049
0.051
0.082
0.070
0.070
0.063
0.061
0.058
0.049
0.051
0.005
0.163
0.138
0.136
0.117
0.116
0.105
0.102
0.096
0.082
0.085
0.007
0.228
0.194
0.190
0.164
0.163
0.147
0.143
0.135
0.114
0.119
0.009
0.293
0.249
0.245
0.211
0.210
0.189
0.183
0.173
0.147
0.153
0.011
0.359
0.304
0.299
0.258
0.256
0.231
0.224
0.212
0.179
0.187
0.013
0.424
0.360
0.353
0.305
0.303
0.273
0.265
0.250
0.212
0.221
The previous table shows that, for a given  , the ACI ratio   c d  is lower than that of EC2 for the
values of fc'  35 MPa which means that ACI provides higher ductility than EC2 in common cases.
The opposite is true for the values of fc'  35 MPa .
But, as the compression concrete strength fc' increases, the difference of   x d   c d  between
the two codes decreases.
EUROCODE 2
ACI318-19
Analysis of Doubly Reinforced Concrete Sections
When compression steel is used, the nominal resisting moment of the beam is assumed to
consist of two parts: the part due to the resistance of the compression concrete and the
balancing tensile reinforcing, and the part due to the nominal moment capacity of the
compression steel and the balancing amount of the additional tensile steel.
The first resisting moment
The first resisting moment
a

M n1  As1 f y  d  
2

a

M n1  0.87 As1 f yk  d  
2

The second resisting moment is that
The second resisting moment is that
produced by the additional tensile and

compressive steel As 2 and
A s'

:
Mn2  A's f s'  d  d'   0.87 A s2 f yk  d  d' 
a

M n  0.87 As1 f yk  d    A's f s'  d  d' 
2

produced by the additional tensile and


compressive steel As 2 and A s' :
Mn2  A's fs'  d  d'   A s2 f y  d  d' 
a

M n  As1 f y  d    A s 2 f y  d  d' 
2



a

øM n  ø  As1 f y  d    A's f s'  d  d'  
2




EUROCODE 2
ACI318-19
Analysis of Doubly Reinforced Concrete Sections
The strain,
 's ,
The strain,  's , in the compression steel is
in the compression steel is
checked to determine whether or not it has
checked to determine whether or not it has
yielded. Initially the stress in the
yielded. Initially the stress in the
compression steel is assumed to be at yield
compression steel is assumed to be at yield

f
f s'
x

 f yd : As f yd   x. cc . fcd .b 
A
A's f yd
;

 A's f yd
 x  d' 
; ε's  ε cu3 

 cc . fcd .b
 x 
s
'
s
c

 f y : As f y  0.85 fc'  1cb  A's f y ;
A
s

 A's f y
0.85
fc'  1b
 c  d' 
;  's  0.003 

 c 
If the strain in the compression steel
If the strain in the compression steel
 's
 's
 y  f yd Es , the assumption is valid and
f s' is at yield, f yd . If  's
 y the compression
 y  f y Es , the assumption is valid and
f s' is at yield, f y . If  's
 y the compression
steel is not yielding, and the value of x
steel is not yielding, and the value of c
calculated above
calculated above
is not correct. A new equilibrium equation
is not correct. A new equilibrium equation
must be written that assumes fs'
must be written that assumes f s'
f yd .
fy .
 x  d' 
As f yd   x. cc . f cd .b  A's 
  cu 3 E s
 x 
 c  d' 
As f y  0.85 fc'  1 cb  A's 
  0.003  E s  c
 c 
x
The value of c determined enables us to
The value of x determined enables us to
compute the strains in both the compression
compute the strains in both the compression
and tensile steels and thus
and tensile steels and thus their stresses
EUROCODE 2
ACI318-19
Analysis of Doubly Reinforced Concrete Sections
Determine the design moment capacity of the beam shown in Figure
the inset of the compression steel is 5cm fck ( EC2)  fc' ( ACI )  20MPa
et f yk ( EC 2)  f y ( ACI )  414MPa
Writing the Equilibrium Equation
Writing the Equilibrium Equation
Assuming
Assuming
f s'  f y
;
As f yd   x. cc . fcd .b  A's f yd
32.16  104  360  0.8  13.33  1  0.35  x
f s'  f y
;
As f y  0.85 fc'  1cb  A's f y
32.16  104  414  0.85  20  0.85  0.35  c
 4.02  104  360
 4.02  104  414
 x  0.27135m
 c  0.23035m
a  x  0.8  0.27135  0.2171m
a  1c  0.85  0.23035  0.1958m
Computing Strains in Compression
Steel to Verify Assumption that It Is
Yielding
Compression steel strain :
 x  d' 
ε's  0.0035 

 x 
Computing Strains in Compression
Steel to Verify Assumption that It Is
Yielding
Compression steel strain :
 c  d' 
ε 's  0.003 

 c 
 0.27135  0.05 
ε 's  0.0035 
  0.00286
0.27135


 0.23035  0.05 
ε's  0.003 
  0.00235
0.23035


ε y  fy Es  414 200000  0.0021
ε y  fyd Es  360 200000  0.0018
ε's  0.00286 ε y  0.0018
ε's  0.00235 ε y  0.0021
Compression steel stress:
 fs'  f y as assumed
 fs'  f y as assumed
A s2 
A s2 
A 's fs'
f yd
4.02  104
360
 360
 4.02  104 m2
A s2 
A 's fs'
f yd
A s2 
4.02  104  414
 4.02  104 m2
414
EUROCODE 2
ACI318-19
Doubly reinforced concrete sections
A s1  A s  A s2 
A s1  A s  A s2 
32.16  104  4.02  104  28.14  104 m2 32.16  104  4.02  104  28.14  104 m2
dx

 x 
s  c 
 0.65  0.27135 
  0.00488
0.27135


d c 

 c 
s  c 
 0.65  0.23035 
  0.0055
0.23035


 s  0.0035 
 s  0.003 
a

M n  As1 f yd  d     A's f s'  d  d' 
2

 s  0.0055
f y Es  0.003  0.0021  0.003
 s  0.0055
0.0051
0.2171 

M n  28.14  10 4  360  0.65 
2 

 4.02  10 4  360  0.65  0.05 
M n  0.6354MNm  635.4kNm
ø  0.9


a

øM n  ø  As1 f y  d    A's f s'  d  d'  
2



øMn  0.6687 MNm  668.7kNm
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Analysis of Doubly Reinforced Concrete Sections
Determine the design moment capacity of the beam shown in Figure
the inset of the compression steel is 5 cm
fck ( EC2)  fc' ( ACI )  25 MPa
et f yk ( EC 2)  f y ( ACI )  500MPa
Writing the Equilibrium Equation
Writing the Equilibrium Equation
Assuming
Assuming
f s'  f y
f s'  f y
As f yd   x. cc . fcd .b  A's f yd
As f y  0.85 fc'  1cb  A's f y
12.56  104  435  0.8  16.67  1  0.35  x
12.56  104  500  0.85  25  0.85  0.35  c
 4.02  104  435
 4.02  104  500
 x  0.0796m
a   x  0.8  0.0796  0.0637m
 c  0.0675m
Computing Strains in Compression
Steel to Verify Assumption that It Is
Yielding
Compression steel strain :
 c  d' 
ε's  0.0035 

 c 
a  1c  0.85  0.0675  0.0574m
 0.0796  0.050 
ε 's  0.0035 
  0.0013
0.0796


Computing Strains in Compression
Steel to Verify Assumption that It Is
Yielding
Compression steel strain :
 c  d' 
 0.0675  0.05 
ε's  0.003 
  0.003 

0.0675


 c 
ε's  0.00078
ε y  f yd Es  435 200000  0.0022
ε y  fy Es  500 200000  0.0025
ε's
ε's  0.00078 ε y  0.0025
 0.0013 ε y  0.0022
Compression steel stress:
 fs'  f yd as assumed
Since the assumption is not valid, we
have to use the equilibrium equation that
is based on fs' not yielding.
 fs'  f y as assumed
Since the assumption is not valid, we
have to use the equilibrium equation that
is based on fs' not yielding.
EUROCODE 2
ACI318-19
Analysis of Doubly Reinforced Concrete Sections
 x  d' 
As f yd   x. cc . f cd .b  A's 
  cu 3 E s
 x 
 c  d' 
As f y  0.85 fc'  1 cb  A's 
  0.003  E s
 c 
12.56  104  435  0.8x  1  16.67  0.35
 c  0.08392m  8.392cm
 x  0.05 
4.02  104 
 0.0035  200000
x


a  1c  0.85  0.08392  0.0713m
 x  0.09017m  9.017cm
a  x  0.8  0.09017  0.0721m
 x  d' 
ε's  0.0035 

 x 
 0.0902  0.05 
ε 's  0.0035 
  0.00156
0.0902


ε's  0.00156
f yd
435

 0.00217
Es 200000
L’acier comprimé n’a pas atteint sa limite
élastique. Compression steel stress :
fs'  ε'sEs  0.00156  200000  312MPa
A s2 
A 's fs' 4.02  104  312

f yd
435
 c  d' 
ε's  0.003 

 c 
 0.08392  0.05 
ε 's  0.003 
  0.001213
0.08392


ε's  0.00121
fy
500

 0.0025
Es 200000
L’acier comprimé n’a pas atteint sa limite
élastique.
Compression steel stress :
fs'  ε'sEs  0.00121  200000  242.5MPa
A s2 
A 's fs'
fy
A s2 
4.02  104  242.5
 0.000195m2
500
A s2  0.000288m2
A s1  A s  A s2
A s1  A s  A s2 
 12.56  104  0.0002  0.00106m2
12.56  104  2.88  104  9.68  104 m2
a

M n  As1 f yd  d    A's f s'  d  d' 
2

0.0721 

M n  9.68  10 4  435  0.45 
2 

 4.02  10 4  312  0.45  0.05 
M n  0.2244MNm  224.4kNm
 0.45  0.08392 
  0.0131
0.08392


 s  0.003 
 s  0.0131
f y E s  0.003  0.0025  0.003
 s  0.0131
0.0055  ø  0.9


a

øM n  ø  As1 f y  d    A's f s'  d  d'  
2




øMn  0.2327 MNm  232.7kNm
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EUROCODE 2
ACI318-19
Design of Doubly Reinforced Concrete Sections
If the moment applied to the section is such that the limit value of x d (c/d) is exceeded
when trying to calculate the section with tension reinforcement only, then compression steel
must be added in order to resist excessive moment and keep the depth of the neutral axis at
the limit value. The configuration for a doubly reinforced beam is given below.
The moment applied to the section Mu is partially balanced by the moment limit
M l determined from the maximum authorized value of x d (c/d) and the rest by the couple
induced by the force of the compression steel and the additional equal force of the tension
steel. We suppose in this calculation that the area of concrete displaced by compression steel
is ignored and that these steel have reached their elastic limit. The strength of compression
steel C s is exactly balanced by an equivalent force of tension steel.
We first calculate the maximum resisting moment of a reinforced section with the maximum
percentage of tensile reinforcement ρ max only.
The corresponding section of tension steel is: A s1  ρ l bd
EUROCODE 2
ACI318-19
Design of Doubly Reinforced Concrete Sections
Residual moment:
Residual moment:
Mres   Mu  Mlim   As2 f yd  d  d' 
Mres  Mu  Mn
Tension steel for : Mres : As2 
Compression steel strain :
 x  d' 
ε's  ε cu3 

 x 
Compression steel stress:
Si ε's  ε y  fs'  0.87fyk
Sinon : fs'  200000  ε's
Tension steel for : Mu  Mres :
Mres
d  d  f
'
yd
A s2 
Mres
ø  d  d ' fy
 ø  0.9 
Compression steel strain :
 c  d' 
ε 's  0.003 

 c 
Compression steel stress:
Si ε's  ε y  fs'  f y
Sinon : fs'  200000  ε's
Compression steel area:
A 's 
A s2 f yd
Compression steel area:
fs'
A 's 
Total tension steel area A s :
As  As1  As2
A s2 f y
fs'
Total tension steel area A s :
As  As1  As2
EUROCODE 2
ACI318-19
Design of Doubly Reinforced Concrete Sections
Design of the following beam : MG  177.5kNm ; MQ  70kNm .
fck ( EC2)  fc' ( ACI )  30MPa
et f yk ( EC 2)  f y ( ACI )  500MPa
Mu  1.35  177.5  1.5  70  344.63kNm
Mu  1.2  177.5  1.6  70  325kNm
fcd  fck 1.5  30 1.5  20MPa
fc'  30MPa  1  0.836
As1  l bd 
As1 
 .max . . fcd
f yd
bd
0.8  0.45  1  30
 0.30  0.40
435  1.5
As1  0.001986m2  19.86cm2
Mlim fcd bd 2  0.295 
As1  l bd 
As1 
Mu  0.345 MNm
Mlim  0.283MNm
 Compression steel needed
Mres  Mu  Mlim
Mres  0.345  0.283  0.062MNm
x  0.45d  0.45  0.40  0.18m
 x  d' 
ε's  0.0035 

 x 
 0.18  0.05 
ε's  0.0035 
  0.00253
0.18


ε's  0.00253
f yd
435

 0.00218
Es 200000
Compression steel reaches its elastic limit
fs
bd
0.836  0.353  0.85  30
 0.30  0.40
500
As1  0.00181m2  18.1cm2
 max 
Mlim 20  0.30  0.402  0.295 
Mlim  0.283MNm
1 .max .0.85. fc'
0.003
fy
Es

 0.006
0.003
 0.353
500
 0.006
200000


max  0.353  0.836  1  0.353
0.836 
 0.252
2 
Ml ACI  0.9  0.85  bd 2 fc'  max
M l ACI 
0.9  0.85  30  0.30  0.40   0.252
2
Ml ACI   0.277 MNm
Mu  0.325 MNm
Ml ACI   0.277 MNm
 Compression steel needed
Mn 1  0.277 0.9  0.308 MNm
Mn  0.325 0.9  0.361 MNm
Mres  Mn  Mn1  0.361  0.308  0.053MNm
EUROCODE 2
ACI318-19
Doubly reinforced concrete sections
Compression steel stress :
 x  d' 
 0.18  0.05 
fs'  700 
 700 


0.18
 x 


fs'
 505.55MPa
 fs'
fyd  435MPa
 fyd  435MPa
Compression steel :
A's

Mres
d  d  f
'
yd
0.062


 0.40  0.05  435
 A s2 f yd  A s2
A s1 f y
0.85fc' b

0.00181  500
 0.118m
0.85  30  0.30
c  a β1  0.118 0.836  0.142m
 0.142  0.05 
 c  d' 
ε 's  0.003 
 0.003 


0.142
 c 


ε's  0.00194
A's  0.000407m2  4.07cm2
A 's fs'
a
A' f '
 s s  A 's
f yd
Tensile steel :
As  As1  As2  19.86  4.07  23.93cm2
ε's  0.00194
fy
500

 0.0025
Es 200000
L’acier comprimé n’a pas atteint sa limite
élastique.
Compression steel stress :
 0.142  0.05 
 c  d' 
fs'  600 
 600 


0.142
 c 


fs'  388.73MPa
500MPa
Compression steel :
A 's 
Mres
0.053

fs'  d  d '  388.73  0.40  0.05 
A 's  0.00039m2  3.90cm2
A 's fs'  A s2 f y  A s2 
A s2 
A 's fs'
fy
0.00039  388.73
 0.00030m2
500
Tension steel :
A s  A s1  A s2  0.00181  0.0003
A s  0.00211m2  21.10cm2
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EUROCODE 2
ACI318-19
Conclusion

The minimum percentage of steel required by the ACI code is greater than that
of the EC2; while the maximum percentage of steel required by EC2 code, to
ensure ductility, is greater than that of ACI ;

In practice the sections in bending require at EC2 addition of compression steel
for moments limit lower than that of the ACI;

The ACI gives smaller steel areas than the EC2 for weakly and heavily reinforced
beams ; this difference being able to reach 15% for values of moments close to
moments limit ;

G = dead load and Q = live load; total load ; P = G + Q . The difference between
the steel areas given by the EC2 and the ACI decreases as Q/P increases.

EC2 is more conservative than ACI concerning the bending resisting moment .

For a given steel percentage, ACI provides higher ductility than EC2 in common
cases; but, as the compression concrete strength increases, the difference in
ductility between the two codes decreases.