Lec.3 working stress 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

24
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

25
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

26
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

27
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

28
  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

29
  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

30
 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 


  MPaffMPa
I
Mc
f crt 5.3257.07.02.14
10906.3
12668010100
9
6




  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.

31
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 
  MPaffMPa
I
Mc
f crt 5.3257.07.022
10463.2
16050010160
9
6




  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





Lec.3 working stress 1

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