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 # 9
28th
Feb 2012
Flexural Analysis and
Design of Beams
(Ultimate Strength Design of Beams)

2. Plain & Reinforced Concrete-1
Ultimate Strength Design of Beams
(Strength Design of Beams)
Strength design method is based on the philosophy of dividing F.O.S.
in such a way that Bigger part is applied on loads and smaller part is
applied on material strength.
fc’
0.85fc’
Stress
Strain
Crushing
Strength
0.003
favg
favg = Area under curve/0.003
If fc’ ≤ 30 MPa
favg = 0.72 fc’
β1 = Average Strength/Crushing Strength
β1 = 0.72fc’ / 0.85 fc’ = 0.85

3. Plain & Reinforced Concrete-1
Ultimate Strength Design of Beams (contd…)
Cc
T = Asfs
la = d – a/2
N.A.
εcu=0.003
Strain Diagram Actual Stress
Diagram
Internal Force
Diagram
In ultimate strength design method the section is always taken as
cracked.
c = Depth of N.A from the extreme compression face at ultimate stage
a = Depth of equivalent rectangular stress diagram.
εs
h
c
d
b 0.85fc
fs
0.85fc
a
Equivalent Stress
Diagram/
Whitney’s Stress
Diagram
a/2
fs

4. Plain & Reinforced Concrete-1
Ultimate Strength Design of Beams (contd…)
Actual Stress
Diagram
0.85fc 0.85fc
Equivalent Stress
Diagram/ Whitney’s
Stress Diagram
c
Cc Cca
• The resultant of concrete compressive
force Cc, acts at the centriod of parabolic
stress diagram.
• Equivalent stress diagram is made in
such a way that it has the same area as
that of actual stress diagram. Thus the Cc,
will remain unchanged.
a/2
ab'0.85fcbf cav ××=××
a'0.85fc'0.72f cc ×=×
c
'0.85f
'0.72f
a
c
c
×=
cβa 1 ×=

5. Plain & Reinforced Concrete-1
Ultimate Strength Design of Beams (contd…)
Factor β1
β1 = 0.85 for fc’ ≤ 28 MPa
Value of β1decreases by 0.05 for every 7 MPa increase in
strength with a minimum of 0.65
0.65'0.00714f1.064β c1 ≥−=
85.0≤

6. Plain & Reinforced Concrete-1
Determination of N.A. Location at Ultimate Condition
CASE-I: Tension Steel is Yielding at Ultimate Condition
ys εε ≥ or ys ff ≥
CASE-II: Tension Steel is Not Yielding at Ultimate Condition
yf
yε sε
yε<sε or ys ff <
0.0015
200,000
300
E
f
ε
y
y === 0.0021
200,000
420
E
f
ε
y
y ===
For 300 grade steel For 420 grade steel

7. Plain & Reinforced Concrete-1
CASE-I: Tension Steel is Yielding at Ultimate
Condition
ysss fAfAT ×=×=
ab'0.85fC cc ××=
2
a
da −=l
abffA cys ××=× '85.0
For longitudinal Equilibrium
T = Cc
bf
fA
a
c
ys
×
×
=
'85.0 1β
a
c =and
Cc
T = Asfs
Internal Force Diagram
a/2
la

8. Plain & Reinforced Concrete-1
CASE-I: Tension Steel is Yielding at Ultimate
Condition (contd…)
Nominal Moment Capacity, Mndepending on steel = T x la






−××=
2
Mn
a
dfA ys
Design Moment Capacity






−××=
2
M bnb
a
dfA ysφφ
Nominal Moment Capacity, Mn depending on Concrete = Cc x la






−×××=
2
a
dab0.85fc'Mn






−×××=
2
a
dab0.85fc'M bnb φφ

9. Plain & Reinforced Concrete-1
Minimum Depth for Deflection Control
I
1
αΔ
3
(Depth)
1
αΔ
For UDL
4
ωLαΔ
( ) 3
LωLαΔ
Deflection Depends upon Span, end conditions, Loads and fy of
steel. For high strength steel deflection is more and more depth is
required.

10. Plain & Reinforced Concrete-1
Minimum Depth for Deflection Control (Contd…)
ACI 318, Table 9.5(a)
Steel Grade Simply
Supported
One End
Continuous
Both End
Continuous
Cantilever
300 L/20 L/23 L/26 L/10
420 L/16 L/18.5 L/21 L/8
520 L/14 L/16 L/18 L/7

11. Concluded