Lec 9-flexural analysis and design of beamns

Plain & Reinforced
Concrete-1
CE3601
Lecture # 9
28th
Feb 2012
Flexural Analysis and
Design of Beams
(Ultimate Strength Design of Beams)

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

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

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 ×=

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≤

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

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

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

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.

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

Lec 9-flexural analysis and design of beamns

Recommended

Recommended

More Related Content

What's hot

What's hot (14)

Similar to Lec 9-flexural analysis and design of beamns

Similar to Lec 9-flexural analysis and design of beamns (20)

More from MUST,Mirpur AJK,Pakistan

More from MUST,Mirpur AJK,Pakistan (20)

Recently uploaded

Recently uploaded (20)

Lec 9-flexural analysis and design of beamns