The document summarizes the working stress design method for reinforced concrete structures. It describes the key assumptions of the method, including that concrete and steel obey Hooke's law, strain is proportional to distance from the neutral axis, and tension in concrete is negligible. The transformed section method is also summarized, where the steel area is replaced by an equivalent concrete area while satisfying compatibility of strains and equilibrium of forces. Several examples are provided to demonstrate calculating stresses in concrete and steel for different beam cross-sections under given loads using the working stress design method.

1. Analysis & Design of
Reinforced Concrete Structures (1) Lecture.3 Working Stress Design Method
23
Dr. Muthanna Adil Najm
1- Working Stress Design Method or elastic method or alternative design method
or allowable stress design method : This method was the principal one used since
1900s to 1960s.
The working stress design method maybe expressed by the following:
Load = service (unfactored) load
aff 
Where:
f = an elastically computed stress, such as by using flexural stress =
I
cM.
for
beams.
fa = a limiting allowable stress prescribed by ACI code , as a percentage of cf  for
concrete and as a percentage of fy for steel.
2- Ultimate Strength Design Method:
In this method, service loads are increased by factors to obtain the load at which
the failure is considered to be " imminent". Also, the section strengths are reduced
by a safety reduction factors.
The ultimate strength design method maybe expressed by the following:
Strength provided ≥ Strength required to carry factored loads
Types of Beams:
1- Types of beams according to section reinforcement:
a- Singly Reinforced concrete beams : main steel reinforcement used at tension
zone only.
b- Doubly reinforced concrete beams: main steel reinforcement used at tension
zone and compression zone.
2- Types of beams according to section Shape:
a- Beams of rectangular section.
b- Beams of ( T, L & I ) section.
Design Methods
b
d
sA'
sA
b
d
sA
h
Singly Reinforced Beam Doubly Reinforced Beam

2. Analysis & Design of
Reinforced Concrete Structures (1) Lecture.3 Working Stress Design Method
24
Dr. Muthanna Adil Najm
c- Beams of irregular sections.
Assumptions:
1- Plain section before bending remains plain after bending.
2- Both concrete and steel obey to Hook's law.


E
3- Strain and stress are proportional to the distance from neutral axis.
4- Concrete strength in tension is negligible.
5- Perfect bond must be maintained between steel and concrete.
6- Allowable stress:
For concrete: cca ff  45.0
For steel : 140saf for fy = 300 and 350 MPa.
170saf for fy = 420 MPa
Structural Behavior of R.C. Beams:
Working Stress Design Methods
b
d
sA
cε
sε
cf
sf
tε tf
cracks
crushing

3. Analysis & Design of
Reinforced Concrete Structures (1) Lecture.3 Working Stress Design Method
25
Dr. Muthanna Adil Najm
Three stages maybe noticed for concrete beam tested in laboratory to failure:
1- Uncracked concrete stage: Full concrete section still works.
2- Cracked concrete stage.
Where : fr = Modulus of rupture of concrete.
3- Ultimate concrete stage.
Transformed Section Method:
From Hook's law:
ccc Ef 
sss Ef 
The basic concept of transformed section is that the section of steel and concrete is
transformed into a homogenous section of concrete by replacing the actual steel
area to an equivalent concrete area.
Two conditions must be satisfies:
1- Compatibility:
b
d
sA
sε
tε
cε
sf
cf 
b
d
sA
cε
sε
ac≤ fcf
as≤ fsf
tε
crt fff  7.0
N.A
kd
cε
tε
ac< fcf
as< fsf
r< ftf
sε
b
d
sA
N.A

4. Analysis & Design of
Reinforced Concrete Structures (1) Lecture.3 Working Stress Design Method
26
Dr. Muthanna Adil Najm
sc   (at the same level: same distance from neutral axis)
c
c
c
E
f
 and
s
s
s
E
f

s
s
c
c
E
f
E
f
  c
c
s
s f
E
E
f 
Or cs nff 
Where : n is the modular ratio and
c
s
E
E
n 
2- Equilibrium:
Force in transformed concrete section = Force in actual steel section
sscc AfAf 
   scsccc nAfAnfAf 
sc nAA 
There are two cases of transformed section:
1- Uncracked Section: where rt ff 
2- Cracked Section: where rt ff 
For doubly reinforced beams, the cracked section is:
b
d
sA
snA
b
kd s= b.kd + nAc)totalA
b
d
sA snA
s1)A-=(nsA-snA
b b

5. Analysis & Design of
Reinforced Concrete Structures (1) Lecture.3 Working Stress Design Method
27
Dr. Muthanna Adil Najm
Ex.1)
For the beam section shown below, if the applied moment is 35 kN.m , fr = 3.1
MPa and n = 9.
1- Calculate the maximum flexural stresses in concrete at top fiber and bottom
fiber and in steel reinforcement.
2- Calculate the cracking moment of the section.
Sol.)
  2
2
1847
4
28
3 mmAs 










    2
1647761847195003001 mmAnbhA s 
Find N.A. location by taking moment of areas about top fiber.
      mmy 265
164776
420184719250500300



        492
33
10513.3265420184719
3
235300
3
265300
mmxI 
1- Flexural stresses:
a- Tension stress at bottom fiber of concrete:
  22
9
6
/1.3/34.2
10513.3
2351035
mmNfmmN
I
Mc
f rt 



Since tension stress at bottom fiber of concrete < modulus of rupture ( rct ff  ),
then section is not crack and hence assumption is true.
b- Compression stress at top fiber of concrete:
s1)A-=(nsA-snA
300 mm
420
mm
3Ø28
500
mm
265 mm
N.A
b
d
s'A
snA
b
kd s1)A'-(2n= b.kd +c)totalA
snA+
sA
s1)A'-(2n

6. Analysis & Design of
Reinforced Concrete Structures (1) Lecture.3 Working Stress Design Method
28
Dr. Muthanna Adil Najm
  2
9
6
/64.2
10513.3
2651035
mmN
I
Mc
fc 



c- Stress in steel:
  2
9
6
/9.13
10513.3
1551035
mmN
I
Mc
nfs 



2- Cracking moment ( Mcr ):
  mkNmmN
c
If
M r
cr .34.46.1034.46
235
10513.31.3 6
9



Ex.2)
Calculate the maximum flexural stresses for the beam section shown below, if the
applied moment is 95 kN.m , and n = 9. Compare with allowable stresses if fy =
420 MPa and f 'c = 25 MPa.
Sol.)
Assume cracked section. Find kd by taking moments about the N.A.
 kddnA
kd
kd s 
2
300   kd
kd
kd  42018479
2
300
kdkd 166236981660150 2

0465441112
 kdkd
   
mmkd 167
2
445111
2
465444111111 2





    492
3
1053.116742018479
3
167300
mmI 
a- Tension stress in concrete:
300 mm
420
mm
3Ø28
500
mm
2
= 9(1847) = 16623mmsnA
kd = 167 mm
N.A
300 mm
420 - kd =
253 mm

7. Analysis & Design of
Reinforced Concrete Structures (1) Lecture.3 Working Stress Design Method
29
Dr. Muthanna Adil Najm
  MPaffMPa
I
Mc
f crt 5.3257.07.07.20
1053.1
1675001095
9
6




 Assumption is true, section is cracked.
b- Compression stress in concrete:
    MPaffMPa
I
Mc
f ccac 25.112545.045.037.10
1053.1
1671095
9
6




.OK
c- Stress in steel:
  MPafMPa
I
Mc
nf sas 1704.141
1053.1
2531095
9 9
6



 .OK
Ex.3)
Calculate the maximum flexural stresses in concrete and steel for the T beam
shown below. M = 100 kN.m , n = 10 and f 'c = 25 MPa.
Sol.)
Assume that N.A. lies within the flange.
  2
14734913 mmAs 
Find kd by taking moments about N.A.
 kd
kd
 600147310
2
900
2
kdkd 147308838000450 2

0196407.322
 kdkd
   
mmmmkd 100125
2
1964047.327.32 2



kd
600-kd
900 mm
nAs
N.A.
100
500
900 mm
3Ø25
250
680

8. Analysis & Design of
Reinforced Concrete Structures (1) Lecture.3 Working Stress Design Method
30
Dr. Muthanna Adil Najm
 Assumption is false and N.A. lies at the web.
Find kd by taking moments about N.A.
     


2
100
10025050100900
kd
kdkd
 kd 600147310
0967046382
 kdkd
   
mmmmkd 100126
2
967044638638 2



       492
33
10906.3126600147310
3
26250900
3
126900
mmI 


a- Tension stress in concrete:
  MPaffMPa
I
Mc
f crt 5.3257.07.02.14
10906.3
12668010100
9
6




 Assumption is true, section is cracked.
b- Compression stress in concrete:
  MPa
I
Mc
fc 23.3
10906.3
12610100
9
6




c- Stress in steel:
  MPa
I
Mc
nfs 35.121
10906.3
12660010100
10 9
6




Ex.4)
Calculate the maximum stresses in concrete and steel for the beam section shown
below. M = 160 kN.m , n = 10 and f 'c = 25 MPa.
Sol.)
2
24637.6154 mmAs 
kd
600-kd
900 mm
nAs
N.A.
250 mm
350 mm
360
70
70
500mm
2Ø28
4Ø28 430-kd
snA
350
kd
s1)A'-(2n
N.A.

9. Analysis & Design of
Reinforced Concrete Structures (1) Lecture.3 Working Stress Design Method
31
Dr. Muthanna Adil Najm
2
12317.6152 mmAs 
Find kd by taking moment about the N.A.
     kddnAdkdAn
kd
kdb ss  12
2
     kdkd
kd
kd  4302463107012311102
2
350
kdkdkd 1416051993551342

0698742752
 kdkd
   
mmkd 160
2
698744275275 2



        4922
3
10463.2160430246310701601231120
3
160350
mmI 
a- Tension stress in concrete:
  MPaffMPa
I
Mc
f crt 5.3257.07.022
10463.2
16050010160
9
6




 Assumption is true, section is cracked.
b- Compression stress in concrete:
  MPa
I
Mc
fc 39.10
10463.2
16010160
9
6




c- Stress in tension steel:
  MPa
I
Mc
nfs 4.175
10463.2
16043010160
10 9
6




d- Stress in compression steel:
  MPa
I
Mc
nfs 93.116
10463.2
7016010160
1022 9
6



