Plain & Reinforced
Concrete

1
CE3601
Lecture # 11 &12
1
st
to 6
th
March 2012
Flexural Analysis and
Design of Beams
(Ultimate Strength Design of Beams)
Plain & Reinforced Concrete

1
Under

Reinforced Failure
C
c
T
Internal Force Diagram
a/2
l
a
Stage

II, Cracked Section
When section cracks, N.A. moves
towards compression face means
“
l
a” increases. “T” and “Cc” also
increase.
Stage

I, Un

cracked Section
N.A. position is fixed, means “
l
a”
remains constant. Only “T” and
“Cc” increase with the increase of
load
Plain & Reinforced Concrete

1
Under

Reinforced Failure
(contd…)
Stage

III, Yielding in Steel Occur
T = A
s
f
y
remains constant and C
c
also
remains constant.
“
l
a” increases as the
N.A. moves towards compression face
because cracking continues.
Failure initiates by the yielding of
steel but final failure is still by
crushing of concrete
C
c
T
Internal Force Diagram
a/2
l
a
Plain & Reinforced Concrete

1
Under

Reinforced Failure
(contd…)
Derivation for ρ
Design Moment Capacity
a
b
n
b
T
M
l
2
a
d
f
A
Φ
M
Φ
y
s
b
n
b
For tension controlled section Φ = 0.9
2
a
d
f
0.9A
M
Φ
y
s
n
b
And
b
'
0.85f
f
A
a
c
y
s
(1)
(2)
Plain & Reinforced Concrete

1
Under

Reinforced Failure
(contd…)
Put value of “a” from (1) to (2)
b
'
0.85f
2
f
A
d
f
0.9A
M
Φ
c
y
s
y
s
n
b
b
'
0.85f
2
f
ρbd
d
f
ρbd
0.9
M
Φ
c
y
y
n
b
For economical design
u
n
b
M
M
Φ
'
0.85f
2
f
ρ
1
f
ρbd
0.9
M
c
y
y
2
u
'
0.85f
f
2
ρ
1
f
0.9
ρ
bd
M
c
y
y
2
u
Plain & Reinforced Concrete

1
Under

Reinforced Failure
(contd…)
Let
(MPa)
R
bd
M
2
u
ω
f
0.85fc'
y
And
Hence
2
ω
ρ
1
f
0.9
ρ
R
y
2
ω
ρ
1
ρ
0.9f
R
y
2
ω
ρ

ρ
0.9f
R
2
y
0
0.9f
R
2
ω
ρ
2
ω

ρ
y
2
0
fc'
fc'
0.85
0.85
0.9f
R
2
ω
ρ
2
ω

ρ
y
2
0
'
0.3825f
R
ω
ρ
2
ω

ρ
c
2
2
2
0.3825fc'
ω
R
4
4
ω
2
ω
ρ
2
2
Plain & Reinforced Concrete

1
Under

Reinforced Failure
(contd…)
By simplification
0.3825fc'
R
1
1
ω
ρ
We have to use
–
ve sign for under reinforced sections. So
fc'
2.614R
1
1
ω
ρ
Reason
For under reinforced section ρ <ρ
b
If we use positive sign
ρ will
become greater than ρ
b
, leading to
brittle failure.
y
1
y
c
b
f
600
600
β
f
'
f
0.85
ρ
< 1.0
ω
ω
ρ
b
Plain & Reinforced Concrete

1
Plotting of R

ρ
ω
d
ρ
ωb
ρbd
b
'
0.85f
f
A
a
c
y
s
2
a
d
ρbdf
Φ
M
y
b
u
2d
a
1
f
ρ
Φ
bd
M
y
b
2
u
2d
a
1
f
ρ
9
.
0
R
y
(3)
(4)
ρ
R
ρ
R
Plain & Reinforced Concrete

1
Trial Method for the determination of “A
s
”
b
'
0.85f
f
A
a
c
y
s
(A)
2
a
d
f
0.9A
M
y
s
u
(B)
2
a
d
0.9f
M
A
y
u
s
(C)
Trial # 1
, Assume some value of
“a” e.g. d/3 or d/4 or any other
reasonable value, and put in
(C)
to get “A
s
”
Trial # 2
, Put the calculated value
of “A
s
” in
(A)
to get “a”. Put this
“a” value in
(C)
to get “A
s
”
Keep on doing the trials unless
“As” from a specific trial
becomes equal to the “As”
calculated from previous trial.
THIS VALUE OF A
S
WILL BE
THE FINAL ANSWER.
Plain & Reinforced Concrete

1
Is The Section is Under

Reinforced or NOT ?
1.
Calculate ρ and if it is less than ρ
max
, section is
under reinforced
2.
Using “a” and “d” calculate ε
t
if it is ≥ 0.005,
section is under

reinforced (tension controlled)
3.
If section is over

reinforced than in the
following equation
–
ve term will appear in the
under

root.
'
f
2.614R
1
1
ω
ρ
c
Plain & Reinforced Concrete

1
Is The Section is Under

Reinforced or NOT ?
(contd…)
1.
For tension controlled section, ε
t
= 0.005
,
d
8
3
β
a
1
Using formula of M
n
from concrete side
a
c
b
n
b
u
C
Φ
M
Φ
M
l
2
a
d
ba
'
0.85f
9
.
0
M
c
u
2
d
8
3
0.85
d
d
8
3
0.85
b
'
0.85f
0.9
M
c
u
2
c
u
bd
'
0.205f
M
b
'
0.205f
M
d
c
u
min
If we keep d > d
min
the resulting section
will be under

reinforced.
d > d
min
means that
section is stronger in
compression.
Plain & Reinforced Concrete

1
Over

Reinforced Failure
C
c
T
Internal Force Diagram
a/2
l
a
Stage

II, Cracked Section
These two stages are same as in
under

reinforced section.
Stage

I, Un

cracked Section
Stage

III, Concrete reaches strain of
0.003 but steel not yielding
We never prefer to design a beam as over

reinforced (compression controlled) as it
will show sudden failure.
Φ = 0.65
ε
s
< ε
y
f
s
<f
y
Plain & Reinforced Concrete

1
Over

Reinforced Failure
Stage

III, Concrete reaches strain of 0.003 but steel not
yielding
(contd…)
2
a

d
ba
'
.85f
0
0.65
M
c
n
b
a
Cc
M
n
b
l
“a” is unknown as “f
s
” is not known
b
'
0.85f
f
A
a
c
s
s
(i)
(ii)
Plain & Reinforced Concrete

1
Over

Reinforced Failure
Stage

III, Concrete reaches strain of 0.003 but steel not yielding
(contd…)
ε
cu
=
0.003
Strain Diagram
ε
s
c
B
C
A
E
D
d

c
Comparing ΔABC & ΔADE
c
c
d
0.003
ε
s
1
1
s
a
d
0.003
ε
a
a
a
β
0.003
ε
1
s
s
s
ε
E
f
a
a
β
0.003
200,000
f
1
s
a
a
β
600
f
1
s
(iii)
(iv)
Eq # (iv) is applicable
when ε
s
< ε
y
Plain & Reinforced Concrete

1
Over

Reinforced Failure
Stage

III, Concrete reaches strain of 0.003 but steel not yielding
(contd…)
Putting value of “f
s
” from (iv) to (ii)
b
'
0.85f
a
a
d
β
600
A
a
c
1
s
(v)
Eq. # (v) is quadratic equation in term of “a”.
Flexural Capacity
2
'
85
.
0
2
a
d
ba
f
a
d
C
M
c
b
c
b
n
b
2
a
d
f
A
Φ
2
a
d
T
Φ
M
Φ
s
s
b
b
n
b
Calculate “a” from (v)
and
“f
s
” from (iv) to
calculate flexural
capacity from these
equations
Plain & Reinforced Concrete

1
Extreme
Tensile
Steel Strain
ε
t
Type of
X

section
c/d
a/d
ρ
max
Φ
<
ε
y
Compression
Controlled
0.65
≥
ε
y
Transition
Section
(Under

Reinforced)
0.65 to
0.9
≥
0.004
Under

Reinforced
(minimum strain
for beams)
0.65 to
0.9
≥
0.005
Tension
Controlled
0.9
≥
0.0075
Redistribution
is allowed
0.9
y
f
600
600
y
1
f
600
600
β
y
y
c
1
f
600
600
f
'
0.85f
β
y
f
600
600
y
1
f
600
600
β
y
y
c
1
f
600
600
f
'
0.85f
β
7
3
7
3
β
1
7
3
f
'
0.85f
β
y
c
1
8
3
8
3
β
1
8
3
f
'
0.85f
β
y
c
1
7
2
7
2
β
1
7
2
f
'
0.85f
β
y
c
1
Plain & Reinforced Concrete

1
Capacity Analysis of Singly Reinforced
Rectangular Beam by Strength Design method
Data
1.
Dimensions, b, h, d and L (span)
2.
f
c
’, f
y
, E
c
, E
s
3.
As
Required
1.
Φ
b
M
n
2.
Load Carrying Capacity
Plain & Reinforced Concrete

1
Capacity Analysis of Singly Reinforced
Rectangular Beam by Strength Design method
(contd…)
Solution
Step # 1Calculte the depth of N.A assuming the section as
under

reinforced
y
s
f
f
y
s
ε
ε
and
b
'
0.85f
f
A
a
c
y
s
1
β
a
c
and
Plain & Reinforced Concrete

1
Capacity Analysis of Singly Reinforced
Rectangular Beam by Strength Design method
(contd…)
Solution
Step # 2 Calculate ε
s
and check the assumption of step# 1
c
c
d
0.003
ε
ε
t
s
For extreme point
If ε
s
≥ ε
y
, the assumption is correct
If ε
s
≤ ε
y
, the section is under

reinforced. So “a” is to be calculated
again by the formula of over reinforced section
b
'
0.85f
a
a
d
β
600
A
a
c
1
s
Plain & Reinforced Concrete

1
Capacity Analysis of Singly Reinforced
Rectangular Beam by Strength Design method
(contd…)
Solution
Step # 3 Decide Φ factor
F
or ε
s
≥ 0.005, Φ = 0.9 (Tension controlled section)
F
or ε
s
≤ ε
y
, Φ = 0.65 (Compression controlled section)
For ε
y
≤ ε
s
≤
0.005, Interpolate value of Φ
(Transition Section)
Step # 4 Calculate
Φ
b
M
n
2
a
d
f
A
M
y
s
b
n
b
2
a
d
ba
'
f
85
.
0
M
c
b
n
b
For under

reinforced Section
For over

reinforced Section
Plain & Reinforced Concrete

1
Capacity Analysis of Singly Reinforced
Rectangular Beam by Strength Design method
(contd…)
Alternate Method
Step # 1 to step # 3 are for deciding whether the section is
tension over reinforced or under

reinforced. Alternatively it can
be done in the following manner
.
1.
Calculate
ρ
and
ρ
max
if
ρ < ρ
max
section is under

reinforced.
2.
Calculate d
min
, if d
≥ d
min
, section is tension
controlled
Plain & Reinforced Concrete

1
Selection of Steel Bars for Beams
1.
When different diameters are selected the maximum
difference can be a gap of one size.
2.
Minimum number of bars must be at least two, one in
each corner.
3.
Always Place the steel symmetrically.
4.
Preferably steel may be placed in a single layer but it is
allowed to use 2 to 3 layers.
5.
Selected sizes should be easily available in market
6.
Small diameter (as far as possible) bars are easy to cut
and bend and place.
Plain & Reinforced Concrete

1
Selection of Steel Bars for Beams
(contd…)
7.
ACI Code Requirements
There must be a minimum clearance between bars
(only exception is
bundled bars).
Concrete must be able to flow through the reinforcement.
Bond strength between concrete and steel must be fully
developed.
Minimum spacing must be lesser of the following
Nominal diameter of bars
25mm in beams & 40mm in columns
1.33 times the maximum size of aggregate used.
We can also give an additional margin of 5 mm.
Plain & Reinforced Concrete

1
Selection of Steel Bars for Beams
(contd…)
8.
A minimum clear gap of 25 mm is to be provided
between different layers of steel
9.
The spacing between bars must not exceed a maximum
value for crack control,
usually applicable for slabs
What is Detailing?
Deciding diameter of bars
Deciding no. of bars
Deciding location of bent

up and curtailment of bars
making sketches of reinforcements.
25mm
Not O.K.
Plain & Reinforced Concrete

1
Concrete Cover to Reinforcement
Measured as clear thickness outside the outer most
steel bar.
Purpose
To prevent corrosion of steel
To improve the bond strength
To improve the fire rating of a building
It reduces the wear of steel and attack of chemicals
specially in factories.
Plain & Reinforced Concrete

1
Concrete Cover to Reinforcement
(contd…)
ACI Code Minimum Clear Cover Requirements
1.
Concrete permanently exposed to earth,
75 mm
2.
Concrete occasionally exposed to earth,
# 19 to # 57 bars
50 mm
# 16 and smaller bars
40 mm
3.
Sheltered Concrete
Slabs and Walls
20 mm
Beams and Columns
40 mm
Plain & Reinforced Concrete

1
Number of Bars in a Single Layer (for beams)
4
.
1
b
02
.
0
N
w
b
Rounded to lower whole number
b
w
= width of web of beam
For I

Shape beam width of bottom flange should be used in
place of b
w
.
Plain & Reinforced Concrete

1
Example
A singly reinforced rectangular beam has a width of 228
mm and effective depth of 450 mm. fc’ = 17.25 MPa, fy =
420 MPa. Calculate flexural capacity for the following three
cases.
1.
2 # 25 bars (SI size)
2.
3 # 25 + 2 # 15 (SI)
3.
Capacity for balanced steel
Concluded
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