Sheryar Bismil Student of Mirpur University of Science & Technology(MUST). Student of Final Year Civil Engineering Department Main campus Mirpur. Here we Gonna to learn about the basic to depth wise study of Plan Reinforced Concrete-i. From basis terminology to wide information about the analysis and design of Concrete member like column,Beam,Slab,etc.

1. Plain & Reinforced
Concrete-1
CE3601
Lecture # 17
3rd
April 2012
Flexural Analysis and
Design of Beams
(Ultimate Strength Design of Beams)

2. Plain & Reinforced Concrete-1
Doubly Reinforced Beams
“Beams having both tension and compression reinforcement
to allow the depth of beam to be lesser than minimum depth
for singly reinforced beam”
 By using lesser depth the lever arm reduces and to develop
the same force more area of steel is required, so solution is
costly.
 Ductility will be increased by providing compression steel.
 Hanger bars can also be used as compression steel reducing
the cost up to certain cost.
 For high rise buildings the extra cost of the shallow deep
beams is offset by saving due to less story height.

3. Plain & Reinforced Concrete-1
Doubly Reinforced Beams (contd…)
 Compression steel may reduce creep and shrinkage of
concrete and thus reducing long term deflection.
 Use of doubly reinforced section has been reduced due
to the Ultimate Strength Design Method, which fully
utilizes concrete compressive strength.
Doubly Reinforced Beam

4. Plain & Reinforced Concrete-1
Behavior Doubly Reinforced Beams
Tension steel always yields in D.R.B.
There are two possible cases:
1. Case-I Compression steel is yielding at
ultimate condition.
2. Case-II Compression steel is NOT yielding
at ultimate condition.

5. Plain & Reinforced Concrete-1
Behavior Doubly Reinforced Beams
Cc
T = Asfs
N.A.
εcu=0.003
Strain Diagram Internal Force
Diagram
εs
h
c
d
b 0.85fc
a
Whitney’s
Stress Diagram
(d-d’)
fs
d
εs’ fs’
Cs
d – a/2
T = Asfs
Cs=As’fs’
Cc=0.85fc’ba
fs=Esεs
fs’=Esεs’

6. Plain & Reinforced Concrete-1
Behavior Doubly Reinforced Beams (contd…)
Case-I Both Tension & Compression steel are yielding at
ultimate condition
fs = fy and fs’=fy
Location of N.A.
Consider equilibrium of forces in longitudinal direction
sc CCT +=
yscys f'Aba'f85.0fA +=
( )
b'f85.0
f'AA
a
c
yss −
=
1β
a
c =and

7. Plain & Reinforced Concrete-1
Case-I Both Tension & Compression steel are yielding at ultimate
condition (contd…)
c
d'c
0.003
'εs −
=
εcu= 0.003
Strain Diagram
εs
c εs’
d’
B
D
E
C
A
Δ ABC & ADE





 −
=
c
d'c
0.003'εs
1
1
s
β
β
c
d'c
0.003'ε ×




 −
=





 −
=
a
d'βa
0.003'ε 1
s
If εs’ ≥ εy compression steel is yielding.
If εs’ < εy compression steel is NOT yielding.
(1)

8. Plain & Reinforced Concrete-1
Case-I Both Tension & Compression steel are yielding at ultimate
condition (contd…)
Cc
T = Asfy
N.A.
Internal Force
Diagram
(d-d’)
Cs
d – a/2
T = total tensile force in the steel
21 TTT +=
T1 is balanced by Cs
T2 is balanced by Cc
s1 CT =
c2 CT =

9. Plain & Reinforced Concrete-1
Case-I Both Tension & Compression steel are yielding at ultimate
condition (contd…)
Cc
T = Asfy
N.A.
Internal Force
Diagram
(d-d’)
Cs
d – a/2
Moment Capacity by Compression Steel
( ) ( )'dd'f'A'ddCM yssn 1
−=−=
( )'ddT1 −=
Moment Capacity by Concrete






−=





−=
2
a
dT
2
a
dCM 2cn2
( ) 





−−=
2
a
dTTM 1n 2
( ) 





−−=
2
a
d'f'AfAM ysysn 2

10. Plain & Reinforced Concrete-1
Case-I Both Tension & Compression steel are yielding at ultimate
condition (contd…)
Total Moment Capacity
21 nnn MMM +=
( ) ( ) 





−−+−=
2
a
d'f'AfA'dd'f'AM ysysysn

11. Plain & Reinforced Concrete-1
Case-II Compression steel is not yielding at ultimate
condition
fs = fy and fs’< fy
'εE'f ss ×=
b'f85.0
'f'AfA
a
c
ssys −
=
1β
a
c =and
a
d'βa
600'f 1
s
−
=
Location of N.A.

12. Concluded