1
INTRODUCTION TO THE
DESIGN
OF
REINFORCED CONCRETE
By
TO
ME
S
ISOP
2
TABLE OF CONTENTS
Chapter I What Reinforced Concrete is, How it Works, and How we Design it.
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5
I

1 Introduction.
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...........................
5
I

2 How We Build.
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......................
8
I

3 How We Design.
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................................
....................
8
I

3.1 Serviceability
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................................
..................
8
I

3.2 Strength
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...........................
8
I

4 Exercises.
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8
Chapter II Materials
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................................
.........................
9
II

1 Unit Stress and Unit Strain
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...
9
II

2 Steel
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................................
......
9
II

2.1 Geometrical Properties of Standard Bars
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................................
......
9
II

2.2
Mechanical Properties: Unit Stress vs. Unit Strain for Static Loads
.....................
9
II

3 Concrete
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................................
................................
9
II

3.1 Tensile Strength
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.............
9
II

3.2 Modulus of Rupture
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.......
9
II

3.3 Compressive Strength
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..
10
II

3.4 Unit Stress vs. Unit Strain (Uniaxial)
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................................
..........
11
II

3.5 Confinement
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................
11
II

3.6 Long Term Response: Shrinkage and Creep
................................
...............................
11
II

4 Exercises
................................
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................................
.............................
11
II

4.1 Axial Load vs. Deformation for a Column with no Reinforcement
............................
11
II

4.2 Axial Load vs. Deformation for a Column with Reinforcement
................................
.
11
Chapter III Flexure
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................................
........................
12
III

1 Introduction
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.......................
12
III

2 Curvature
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...........................
12
III

2.1 Definition
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...................
13
III

2.2 The Relationship between Curvature and Deflection
................................
.................
15
III

3 The Relationship b
etween Curvature and Bending Moment
................................
............
18
III

4 Stages of Response
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............
20
III

5 Linear Response (Stages I and II) and Serviceability
................................
.......................
21
III

5.1 Response before Cracking (Stage I)
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................................
...........
21
III

5.2
Response after Cracking (Stage II)
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................................
.....
23
III

5.3 Limiting Crack Width
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................................
................................
27
III

5.4 Limiting Immediate Deflections
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................................
................
27
III

5.5 Limiting Long

Te
rm Deflections
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...............
27
III

6 Nonlinear Response (Stage III) and Strength
................................
................................
....
28
III

6.1 Beams with a Single Layer of Reinforcement
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...........................
28
III

6.2 Comparison between Stages I, II, and II
................................
................................
....
33
III

6.3 Balanced Failure
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.........
33
III

6.4 Beams with Compression Reinforcement
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................................
..
33
III

6.5 Reinforcement Limits
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................................
................................
.
33
III

6.6 T

Beams
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................................
.....................
33
III

7 Design of Elements to Resist Flexure
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33
III

8 Examples
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...........................
33
III

8.1 Simple Be
am Design
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................................
..
34
III

8.2 Deflection of a Beam in a Frame
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...............
34
III

8.3 One Way Slab
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............
34
III

9 Interactive Exercises
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.........
34
3
III

10 Exercises
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.........................
34
III

10.1 Design, Construction, and Test of a Small

Scale Beam
................................
...........
34
Chapter IV Bending and Axial Load: Columns
................................
................................
............
35
IV

1 The Basics: Columns without Longitudinal Reinforcement
................................
.............
35
IV

2 Columns with Longitudinal Reinforcement
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......
35
IV

3 The Idea Behind Prestressed Concrete
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35
IV

4 Examples
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...........................
35
IV

5 Interactive Exercises
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.........
35
IV

6 Exercises
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...........................
35
Chapter V How Concrete and Steel Interact: Bond
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.......
36
V

1 Anchorage and Development
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36
V

2 Lap Splices
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36
V

3 Reinforcement Layout
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........
36
V

4 Examples
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................................
............................
36
V

5 Interactive Exercises
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................................
................................
..........
36
V

6 V

1
Exercises
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..................
36
Chapter VI Shear
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...........................
37
VI

1 The Initial State o
f Stresses and Mohr’s Circle
................................
................................
37
VI

1.1
37
VI

2 The Facts: Test Results
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.....
37
VI

3 The Myth: The Imaginary Truss
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37
VI

4 The Fable of Vc and Vs
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....
37
VI

5 The Effect of Axial Load
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................................
..
37
VI

6 Examples
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................................
................................
...........................
37
VI

7 Interactive Exercises
................................
................................
................................
.........
37
VI

8
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................................
...........
37
VI

9 Exercises
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...........................
37
4
LIST OF FIGURES
5
Chapter I
What Reinforced Concrete is
,
How it
Works
, and How we Design it.
I

1
Introduction.
The marriage between concrete and steel
has lasted
long
.
T
he fir
st structures we know of in
which stone and
iron
were used together were built by the ancient Greeks
(
Figure
I

1
)
.
Over the
years, the stone was replaced
by
concrete and iron
by
steel.
The
reasons for the union between
steel and
concrete are simple. C
oncrete is strong in compression but
we
a
k
in tension.
In contrast,
s
teel
bar
s
are
strong in
tension
.
If we put
steel
bars
into concrete
,
we get a composite
that can
handle both compression
and
tension
.
And w
e can use
this composit
e
to make elements of
almost
any shape we can imagine
without
the need for
melting
and handling molten material at high
temperature
.
We shape concrete by casting it in
to
molds soon after we mix its ingredients
–
stone,
sand, water and cement
–
while the mix
is still a fluid. The operation is relatively simple.
In
addition, the composite is lighter and,
in many instances
, cheaper than steel.
Putting
concrete and
steel together
,
we gain the best of both
materials
.
Figure
I

1
Greek Structure, Western Turkey
Reinforcing steel comes in the shape of bars
with circular cross sections
.
It has been said that one
must place reinforc
ing bars
in concrete anywhere one can imagine the concr
ete may crack.
Cracks in concrete
form
in different directions. So we
use
a “mesh,”
or
a cage of steel
reinforcement embedded
in the concrete (
Figure
I

2
).
In this cage, we call the bars parallel to the
longitudinal axis of the
element “longitudinal reinforcement.” Bars in the perpendicular direction
are called “transverse reinforcement.”
6
L
o
n
g
i
t
u
d
i
n
a
l
R
e
i
n
f
o
r
c
e
m
e
n
t
T
r
a
n
s
v
e
r
s
e
R
e
i
n
f
o
r
c
e
m
e
n
t
Figure
I

2
Reinforcement
To understand the function of reinforcement
in concrete
,
thin
k of a prismatic beam
, a beam of
constant cross section,
supported by a roller and a pin (
Figure
I

3
)
.
Imagine that a
concentrated
load is applied at
midspan
gradually
. If the beam is not reinforced, it fails when the applied lo
ad
causes the first crack to appear (at the section with maximum bending moment). The crack is the
result of failure of the concrete subject
ed to normal tensile
unit stress
es.
M
u
l
t
i
p
l
e
n
a
r
r
o
w
c
r
a
c
k
s
w
i
d
e
c
r
a
c
k
R
e
i
n
f
o
r
c
i
n
g
B
a
r
s
Figure
I

3
Simple Beam
If w
e plot the magnitude of the applied load versus maximum deflection, we obtain the curve in
Figure
I

4
(
solid line
).
Observe that the load drops suddenly with the appearance of the crack
,
which
takes place
with no visible warning
signs
preceding it
(Point A)
.
If we place
reinforcing
bar
s near
the
bottom of the beam, things
change radically
(dashed line)
. The addition
of the bar
s
does not
lead to
crack prevention
.
With loading,
the
c
oncrete
in the beam
still cracks
and it does
so
at about the same load that caused cracking in the beam with no reinforcement
. B
ut the
reinforcement
allows the
beam
to
carry
larger
loads
and
to
reach deflections
several
times larger
than the deflection at first cracking (Point B)
.
7
Figure
I

4
Load

Deflection Relationship for a Beam with and without Reinforcement
The implications of the
phenomena
described
are tremendous.
Before additi
on of reinforcement,
the beam i
s weak a
nd brittle. We add
reinforcing bars
(usually made out of scrap steel),
and we
ge
t a beam that can be several times stronger and is
also
ductile
,
that is, it reaches large
deformations
–
which serve as warning
–
before failure.
The
benefits of the addition
of reinforcement have limits. Too much or too little reinforcement
lead to
far less desirable responses. Too little
reinforcement
may lead to a brittle failure that one
cannot distinguish from the failure of plain concrete. Too much reinforcement can
al
so
lead to
brittle failure of the concrete in compression or in shear
.
S
hear failure, accounts for a
large
fraction of the
less lustrous entries
in
the record of
reinforced concrete.
Figure
I

5
shows
the
photograph of
a column in
a bridge after the 1995 Kobe
E
arthquake
.
The damage, obvious and
impressive, was caused by shear induced by inertial forces in the deck of the bridge.
Figure
I

5
Reinforced Concrete Column
Before and A
fter Kobe (1995) Earthquake
Lack of bond between the reinforcement and the surrounding concrete may also lead to trouble.
Imagine
,
for instance
,
that the bars
in the beam we described before
are covered with
grease
before concrete is cast around
t
hem
.
Y
ou must admit that to expect those bars to be of help
would be to expect too much. The bars can only be of help if the concrete can transfer forces to
them
so that the reinforcement can resist tensile forces where the concrete has cracked.
That
Plain Concrete
D
e
f
l
e
c
t
i
o
n
L
o
a
d
C
r
a
c
k
i
n
g
A
B
8
transfer
is not going to
be
efficient if there is
grease
in
between the concrete and the
reinforcement.
The lack of bond would render
the reinforcement useless.
I

2
H
ow W
e Build.
Figure
I

6
Structural System
I

3
H
ow W
e Design.
Service
Ultimate
I

3.1
Serviceability

Control of crack widths

Control of deflections
I

3.2
Strength
I

4
Exercises
.
1.
2.
3.
4.
9
Chapter II
Material
s
II

1
Unit Stress and Unit Strain
II

2
Steel
II

2.1
Geometrical Properties of Standard Bars
II

2.2
Mechanical Properties:
Unit
Stress vs.
Unit
Strain for Static
Loads
Figure XXX shows a representative stress

strain curve for reinforcement.
II

3
Concrete
II

3.1
Tensile Strength
II

3.2
Modulus of Rupture
The modulus of rupture is related to but not
the same as the tensile strength of the concrete. In
effect, the modulus of rupture is an abstract quantity determined from bending test of a concrete
beam. At cracking, the distribution of tensile
unit
stress over the cross

section of the beam is not
0
10000
20000
30000
40000
50000
60000
70000
80000
90000
100000
0
0.05
0.1
0.15
0.2
Unit Strain
Unit Tensile Stress, psi
s
h
y
10
lin
ear (
Figure
II

1
) but the modulus of rupture is calculated assuming that it is linear. The
modulus of rupture is therefore expected to be approximately 50 % higher than the tensile
strength. The error would be compensated in appli
cations to beams with rectangular sections. But
we need not be preoccupied with this error because there are many other factors that affect the
actual tensile strength. For example, presence of longitudinal reinforcement would cause the
concrete to incur t
ensile
unit stress
es if the concrete shrank, and this
unit stress
would reduce the
effective tensile strength. What we need to remember is that the calculated cracking moment is
always an approximate quantity.
Figure
II

1
Unit
Strain and
Unit
Stress
Distributions at Cracking
In
reinforced concrete design in the
US, it is customary to assume the
modulus of rupture
for
normal weight concrete to be related to the compressive str
ength
f
r
7.5
f'
c
II

1
with the understanding that f
c
is in psi and the value of
c
f
is also in psi.
Keep in mind that the modulus of rupture can be smaller than indicated by
Eq.
II

1
. Eq.
II

1
is a
reference value; a convention. A reasonable lower bound to the modulus of rupture is closer to
c
'
f
6
.
II

3.3
Compressive Strength
Cylinder
Strength
Figure 7a
shows a measured stress

strain curve for a concrete cylinder
in compression. The peak stress is 4
ksi and we refer to it as f
c
’. We note
that, up to a stress of approximately
0.5f
c
’ (2 ksi for the case considered),
the slope of the curve is nearly
constan
t. It can be represented by a
straight line. We shall therefore
limit the use of expressions 15 and
16 to cases in
which the
maximum stress in the concrete
does not exceed 1/2 the peak stress.
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
0.000
0.001
0.002
0.003
0.004
0.005
0.006
0.007
Unit Strain
Unit Stress, psi
ksi
11
Beyond that stress, stiffness decreases with increase in strain
(concrete softens) until a peak value
is reached at a compressive strain of approximately 0.002. As strain increases beyond that value,
stress tends to decrease. The rate of decrease depends on the quality and strength of the concrete
as well as the rate
and method of load application.
Ritter defined the initial portion of the stress

strain curve for concrete by a parabola known as the
Ritter Parabola
. Hognestad
recognized the limited plasticity in concrete by extending the curve
defined by Ritter, with a
straight line having a negative slope, to a useful limit of compressive
strain which he set at 0.0038 on the basis of observations from a series of tests of reinforced
concrete elements subjected to axial load and bending. Experimental studies made since
Hognestad’s work have shown that the strain limit may vary depending on the strain gradients
across the section and along the element and that there is seldom reason to quote it in two figures.
In ACI 318

08, this value is rather arbitrarily set at 0.003.
Higher values are plausible, however
they are “illegal” for design.
Strength of Concrete in a Column vs Cylinder Strength
II

3.4
Unit
Stress vs.
Unit
Strain
(Uniaxial)
Hognestad
Whitney
II

3.5
C
onfinement
II

3.6
Long Term Response:
Shrinkage
and Creep
II

4
Exerc
ises
II

4.1
Axial Load vs. Deformation for a Column with no Reinforcement
II

4.2
Axial Load vs. Deformation for a Column with Reinforcement
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx SP 05/28/08
12
Chapter III
Flexure
III

1
Introduction
An element in a structure can b
e subjected to a number of mechanical actions including bending
moment, axial load, shear, and torsion. In reinforced concrete, the effects of bending and axial
load are better understood that the effects of shear and torsion. We, therefore, introduce th
e basic
concepts related to bending and axial load first.
We consider the behavior of a prismatic reinforced concrete element subjected to
increasing
bending moment
, as shown ideally in
Figure
III

1
. Please note
that
moment is
c
onstant along the segment shown and,
therefore, there is no shear. We ignore
the self

weight of the element.
As the beam is subjected to moment
it
bends
(
top fibers shorten and bottom
fibers elongate
)
. How much
the beam
bends
is related to the magnitude
of the
bending moment
, the proportions of the
cross section, and the properties of the
materials
.
We study this relationship in
this chapter.
III

2
Curvature
We quantify bending using a quantity
called curvature. Although curvature is
a concept that is mor
e abstract than
other measures of deformation such as
deflection and slope, it a fundamental
parameter
in the
design reinforced
concrete elements to resist flexure. The
student is therefore encouraged to
follow this section with attention.
Figure
III

1
Beam Subjected to End Moments
13
III

2.1
Defin
ition
Consider the beam
shown in
Figure
III

2
. The beam supports loads
between two supports. These loads
cause the beam to deflect. We
idealize the deflected shape using
two hinges and three links. The
deflections of the idea
lized and
“actual” beams coincide at the
locations of the hinges but not
anywhere else
.
As the idealized beam deforms,
each hinge rotates. To
accommodate this rotation, fibers
above the hinge compress and fibers
below the hinge stretch
(or fracture)
.
Th
e fiber passing through the center
of the hinge does not compress or
stretch.
We call this
fiber
“neutral
axis.”
The fiber that deforms the
most is that which is farthest from
the
neutral axis
.
We assume that
the amount of deformation in each
fiber is di
rectly proportional to the
distance to the neutral axis
(
Figure
III

3
)
:
n
i
y
III

1
Dividing deformation by initial
length we obtain unit strain:
x
y
n
i
III

2
or
n
y
III

3
where
=
/
x is
the
average
rotation per unit of length
and i
t is
a
measure of the
amount of fle
xural
deformation
in a segment of the beam
.
This quantity
is called curvature
and,
a
s
x tends to zero
(i.e. as we
Figure
III

3
Definition of Curvature
C
x/2 = L/4
1
3L/4
2
L/4
A
B
C
1
2
x/2
x/2
x/2
1
2
0
A
B
A
C
Load
Distance
Deflection
Distance
Approx. Deflection
Distance
Hinge
Rigid Link
2
1
t
B/A
B
B
Figure
III

2
Idealization of a Deformed Beam.
Fiber
A
B
x/2
x/2
i
n
y
y
d
14
subdi
vide the beam in more segments
),
curvature approaches the second
derivative of the function describing
the de
flected shape of the beam.
The
second derivative of the deflection is
the rate of change in the slope
.
Recall
from calculus that functions with a
positive second derivative are concave,
that is, shaped as a bowl (
Error!
Reference source not found.
).
Functions with a negative second derivative are convex.
If we use
the
dist
ance
to the extreme fiber in compression
d
y
, instead
of distance to the neutral
axis
as reference
, and assigning a negative sign to tensile
unit strain
, we obtain
:
y
c
y
d
)
d
(
III

4
where
c
is the maximum compressive strain.
Note that
( ) 0
y
d
at the
neutral axis
.
The
distance
c
from the extreme fiber in compression to the neutral axis is:
c
c
III

5
And, therefore,
c
c
III

6
Expressing
in terms of the tensile strain
s
at a depth
d
from the outermost fiber in
compression:
c
d
s
III

7
T
he
expressions
derived in this
section
describe the relationship between curvature, strain
, and
neutral axis depth
. They
are purely geometric and do not includ
e terms related to the applied
loads.
If our idealizations (
Figure
III

2
) seem a bit too far from reality consider the photographs in
Figure
III

1
. Notice that the presence of cracks causes
increases in
rotation
and deformation
at
discrete
locations
along the beam.
This observation does not justify our idealization but it does
make us realize that we are not dealing with a continuum and
that when we compute
unit
strain
s
in
reinforced
concrete we
shall c
onsider our results
average
values
.
Remember
:
Curvature is a measure of the amount of deformation caused by bending. It is
defined as the rate of change in the slope with distance along the axis of the beam
(Eq
.
III

3
) and is equal to the ratio of maximum compressive strain to depth to
neutral axis (Eq
s
.
III

6
).
x
y
a
x
2
y
2
4
dx
y
d
2
2
b
x
y
2
2
dx
y
d
2
2
2
2
1
x
y
1
dx
y
d
2
2
Figure
III

4
Second Derivative
15
III

2.2
The Relationship
b
etween Curvature and
Deflection
There is a unique relationship
between curvature and deflection.
We study this relationship
using the example in
Figure
III

2
. For
small deformations, the
“
deviation
”
of
point
C with respect to the tangent at A is
L
4
1
L
4
3
t
2
1
A
/
C
L is the length of the beam. T
he slope
at
A
can be obtained dividing this result by L:
4
1
4
3
2
1
0
Similarly, the deviation of B (
midsp
an)
with respect to
the tangent at
A is:
L
4
1
t
1
A
/
B
and, therefore, t
he deflection at B
can be estimated as
8
L
L
4
1
2
L
4
1
4
3
t
2
L
2
1
1
2
1
A
/
B
0
B
Recalling that rotation and curvature are related
to one another (
=
x
, with
x=L/2 for this
case
)
16
L
2
2
1
B
This
expression
is
an
approximation.
F
or a parabolic distribution of curvature
(such as the
curvature distribution in a linear, prismatic beam under uniform load
–
Figure
III

5
)
with
maximum curvature at midspan
m
it
yields
2
m
2
m
B
L
32
3
16
L
4
3
4
3
We know the answer should be 5/48
m
L
2
. Our estimate is wrong. But it is not wrong by much.
The error, 10%, is not too large considering the simplicity of the method
used. For a triangular
distribution of curvature, the error is larger (the student is asked to compute it).
7
16
m
3
4
m
15
16
m
m
A
B
C
x/2 = L/8
Figure
III

5
Parabolic Distribution of Curvature
16
What is of importance is not the accuracy of our expressions
–
we know they are not accurate
–
but
that they
demonstrate that
there is a unique re
lationship between curvature distribution and
deflected shape
. If one is given, the other can be computed.
To improve our estimate of
deflection we
divide the beam in
more segments
as shown in
Figure
III

6
.
The rotation at eac
h hinge
contributes to the deviation from
point C to the tangent at A. The
contribution from each hinge is the
product of its rotation and the
distance to C:
n
1
i
2
1
i
C
/
A
x
i
n
t
III

8
n is the total
number of hinges.
Replacing
with
x:
n
1
i
2
1
i
A
/
C
x
i
n
x
t
III

9
The first term in this
summation
represents an area under a diagram of
the distribution of curvature. The
second term is
the
distance
from each
hinge
to point C.
We conclude that
the deviation of C with respect to the
tangent at A is equal to the moment
with respect to C of the area between
A and C.
Similarly, the deviation of
B with respect to the tangent at A is
:
n
2
1
1
i
2
1
2
1
i
A
/
B
x
i
n
x
t
III

10
And t
he deflection at B is:
A
/
B
A
/
C
B
t
2
t
III

11
x
x
x
x
x
x
x
x
x/2
Hinge
Rigid Link
1
2
3
4
5
6
7
8
4
C/A
4
9
x/2
t
t
B/A
B
B
A
C
A
C
B
4
A
C
Curvature
Distance
Deflection
Distance
Figure
III

6
The Relationship be
twee
n Curvature and
Deflection
17
If we use four hinges
(n=4,
x=L/4)
, and assuming again a parabolic distribution of curvature
centered at midspan
(
m
16
7
4
1
,
m
16
15
3
2
–
Figure
III

5
) we obtain
:
2
m
32
11
2
m
16
1
2
1
16
7
2
3
16
15
2
5
16
15
2
7
16
7
A
/
C
L
L
t
,
2
m
128
9
2
m
16
1
2
1
16
15
2
3
16
7
A
/
B
L
L
t
and
2
m
128
13
A
/
B
A
/
C
2
1
B
L
t
t
which
does not deviate more than 2.5% from
the correct answer (5/48
m
L
2
).
The same result
can be obtained computing the deviation of A with respect to the tangent at B:
2
m
128
13
2
m
16
1
2
3
16
15
2
1
16
7
B
/
A
L
L
t
This is no coincidence. It is a consequence o
f the fact
that, for a distribution of curvature
that is
symmetric with respect to midspan
,
the tangent
to the deflected shape of the beam
at midspan
is
horizontal.
Notice that
t
A/B
and t
B
/
A
are different. And notice that t
A/B
represents an approximate
e
stimate of
the moment with respect to A of the shaded area in
Figure
III

5
. The
shaded area is
m
2
1
3
2
B
/
A
L
This area represents the total change in slope from A to B
(which is equal to the slope at A)
.
Its
moment w
ith respect to A is
the deflection at midspan:
2
m
48
5
m
2
1
3
2
2
1
8
5
B
/
A
L
L
L
t
This result shows us that we do not always have to use the analogy of the links connected by
hinges. But the analogy helps understand the problem and is useful in problems where the
distribution of curvature does not have simple geometry.
18
x
C
=
T
T
F
F
j
.
d
F
III

3
The Relationship between Curvature
and Bending Moment
W
e have seen that deformations caused by bending can be quantified in terms of
curvature
, and
that
curvature is related directly
strain (Sec
tion
III

2
)
.
In
design
,
it is important
to relate strains to
bending moment
because we judge whether an element is approaching failure by comparing
computed strains with strains observed to be associated with failure in uniaxi
al tests (
Chapter II
)
.
So we need to be able to relate moment
to curvature
in order to
be able to relate
moment
to
strain
(
Figure
III

7
)
.
Figure
III

7
Curvature, Strain, Moment
To understand how a beam resists bending
moment and how the mechanism of
resistance is related to curvature, consider
first the craw bar shown in
Figure
III

8
.
The craw bar
is used to pull a nail from a
wall. Force F causes a moment
of
magnitude F
∙
x
with respect to the face of
the wall. That moment is counteracted by
a
force
couple: T and C. C causes
compression in the bar. T causes tension
in the nail, and balance
s C in the
horizontal direction
:
T
C
III

12
Equilibrium of moments requires:
d
j
T
x
F
M
III

13
Strain
Moment
Curvature
Sec.
III

3
Sec.
III

2
Design
(Sec.
III

6
)
Figure
III

8
Craw Bar
Analogy
19
Consider
now
the beam in
Figure
III

9
.
Consider the
equilibrium of the segment
t
o
the left
of section X
.
I
nternal forces must balance the moment caused by the
reaction
with respect to X.
Internal forces are
distributed through the cross section.
We know that
fibers below the neutral axis stretch and fibers above
shorten. We infer t
hat b
elow the neutral axis
there are
tensile
unit
stress
es
. Above
it
,
there must be
compressive
unit stress
es
. We think of the resultants
of these stresses as
two
forces
C
(compression)
and T
(tension)
, analogous to
the forces
acting on the craw
bar
.
Aga
in
, equilibrium of
forces and
moment
s
require
:
T
C
III

14
a
nd
d
j
T
M
III

15
The
magnitude of the resultants C and T
that balance
the moment M
and the distance between them (jd)
depend on the
distribution of
unit stress
es
(as it is
explained in Sections
III

5
and
III

6
)
.
The distance jd
is called “internal lever
arm” and is essential in the
design of beams.
Equation
III

15
describes how moment is related
to the resultants of internal normal
unit stress
es.
These stresses are associated with
normal
strain
s
(
Chapter II
)
.
Recalling that, in turn,
strain
is
related to curvature (Eq.
III

4
)
,
w
e conclude that
there is a relationship between
applied
m
oment
and curvature
.
This
relationship can be illustrated
using
again
the analogy of the
craw bar
as shown in
Figure
III

10
.
In this figure, the mechanism
resisting tensile forces is represented by a spring.
The a
pplied force creates a moment. This
m
oment cause
s
the bar to rotate.
And the r
otation of the bar cause
s the spring to stretch. This
stretch is associated with a force in the spring. This force and the compressive reaction C
form
the
force couple that
balances the applied moment. Moment, bar rotation and spring deformation
are
,
therefore
,
related to one
another.
In the actual beam, c
urvature is analogous to the
amount of rotation of the craw bar
(curvature is
rotation per unit length)
.
And
tensile strain is
analogous to
the stretch in the spring
(strain is
deformation per unit length)
. Moment, curvatur
e,
and strain are thus related to one another
in the same
way
that
moment, bar rotation, and spring stretch are
related
.
Next we study how the relationship between
moment, curvature, and strain
changes depending on
the magnitude of the applied moment and
the
properties of the beam
.
Figure
III

9
Internal Forces Associated
with
Bending
C
C
Figure
III

10
P
P
/
2
P
/
2
P
/
2
=
V
C
(
T
)
T
(
a
)
(
b
)
X
(
d
)
T
e
n
s
i
o
n
X
(
c
)
M
=
T
.
j
.
d
=
V
.
x
j
.
d
V
x
C
o
m
p
r
e
s
s
i
o
n
20
III

4
Stages of Response
To
describe the relationship between moment,
curvature and strain quantitatively
, we first
consider an element with a single layer of
reinforcement. If this element is subjected to
bending as shown in
Figure
III

1
, it may go
through four different stages of response.
Initially (M<M
cr
in
Figure
III

12
), the element
responds linearly (an increase in moment is
directly proportional to an increase in
displace
ment). At M
cr
, the concrete in the tension
face
is assumed to crack.
Beyond that point,
displacement increases more rapidly with increase
in moment.
At M
y
, the reinforcement yields. There is a drastic
change in stiffness. Displacement increases with v
irtually no change in moment. Depending on
the amount of tensile reinforcement, the moment may start increasing again at a large
displacement as shown, or it may start decreasing as indicated by the broken curve in
Figure
III

12
.
OVER

REINFORCED BEAMS?
Figure
III

12
Typical Moment

Deflection Relationship
for Beam Subjected to Bending
T
he relationship between moment
, curvature,
and strain
is different i
n each stage
of response.
To
consider each stage separately,
we
shall call the range from zero moment to M
cr
Stage I. The
range from M
cr
to M
y
will be called Stage II, and the range beyond M
y
will be called Stage III. In
the design mode, we are seldom int
erested in Stage IV, the range where the moment may start
increasing again. We shall focus on the first three stages.
Stage I
Stage II
Stage III
M
cr
M
y
If strain hardening occurs in reinforcement
If concrete fails (crushes) in compression
Initiation of Stage IV
Yield
Crackin
g
Moment
Deflection
Figure
III

11
Cross Section of
Beam with Single Layer of
Reinforcement
d
b
Tensile
Reinforcement
Area = A
s
h
21
III

5
Linear Response
(Stages I and II)
and Serviceability
In Section
I

3
we discussed that effective design re
quires consideration of the response of the
reinforced concrete element to different levels of load.
Unit s
tresses caused by service

level
loads
–
loads related to daily use of the structure
–
should remain within the linear ranges of
response for concrete
(in compression) and steel. Service

level
loads should not cause
unit
stress
es or strains approaching those associated with failure of the materials.
It is important
that
we understand the relationships between service

level loads and
the
unit
stress
es
an
d strains
they
produce
because these relationships are useful in the development of methods to control crack
width
s
and deflections.
III

5.1
Response before Cracking
(Stage I)
d
b
h
f
h/2
h/2
c
c
(a) cross section
(b) unit strain
(c) unit stress
Assuming that the section
(
Figure
III

11
)
is uncracked and that the effect of the reinforcement on
section stiffness is small
(which it is known from theory and from observation)
, the necessary
theory requires two assumptions:
1.
Distribution of unit strain ov
er depth of section is linear (
Figure
III

13
b
).
2.
The normal
unit
stress is linearly related to
unit
strain in tension and compression
(
Figure
III

13
c
)
by
the
Young’s Modulus
for Concrete, E
c

obtained fr
om tests of 6x12

in.
cylinders subjected to uniaxial compression
(Sec.
II

3.4
)
.
Because we have assumed a
constant modulus in tension and
compression, we recognize that the
neutral axis (position with zero
applied stress) w
ill be, according to
our idealization, at mid

height of the
rectangular section.
Realizing that
the conditions described by
Equations
III

14
and
III

15
apply in
all the stages of response, t
he
internal fo
rces and resisting moment
Figure
III

13
f
c
b
h
/
2
h
/
2
h
/
3
h
/
3
C
T
Figure
III

14
Unit Stress Distribution before Cracking
22
can be expressed i
n te
rms of the extreme

fiber
unit
stress
f as
(
Figure
III

14
)
:
c
c
f
h
b
4
1
2
h
b
f
2
1
T
C
III

16
c
2
f
h
b
6
1
h
3
2
T
M
III

17
The factor 2/3h in Equation
III

17
is the distance between the resultants of compressive and
tensile stresses
; the internal lever arm jd:
h
3
2
d
j
III

18
This
result
is a
consequence
of the assumed distribution of
unit
stress.
Unit stress distributions
with different shape lead to different results.
The maximum compressive strain is
2
6
1
c
c
c
max
bh
E
M
E
f
III

19
And c
urvature
is
g
c
c
I
E
M
c
III

20
where
3
12
1
g
bh
I
and c = h/2
.
From mechanics of materials, we know the
relationship
described by Eq.
III

20
to be true.
The boundary between Stages I and II is given by the moment at cracking M
cr
.
The moment at
cracking
can
be expressed as
c
2
r
2
cr
'
f
4
5
h
b
f
h
b
6
1
M
III

21
Here f
r
is
modulus of rupture (Sec.
II

3.2
).
Remember
:
Before cracking
–
in Stage I
–
the relationship between curvature and bending
moment is:
g
c
c
I
E
M
c
It is customary to assume that response remains in Stage I as long as the applied
moment does n
ot exceed:
c
2
r
2
cr
'
f
4
5
h
b
f
h
b
6
1
M
23
III

5.2
Response after Cracking
(Stage II)
(a) section
(b) unit strain
(c) concrete
(d) resultants
unit stress
Stage two starts when the applied moment first exceeds the moment at cracking
(
Equation
III

21
).
I
n stage II, we ignore tensile
unit
stress
es
in the concrete but we retain the assumptions we made
above for the strain distribution and the relationship between compressive
unit
strain and
unit
stress in the
concrete.
As illustrated in
Figure
III

15
, the depth to the neutral axis is expressed as kd where d is the depth
to the centroid of the tensile reinforcement from the extreme fiber in compression. From
the
geometry of the assumed
distribution of
strain (
Figure
III

15
), we obtain the relationship
s
c
1
k
k
III

22
s
= unit strain in reinforcement
c
= unit stra
in in concrete at the extreme fiber in compression
k = ratio of neutral

axis depth from extreme fiber in compression to the effective depth of the
reinforcement
From equilibrium of
normal
forces acting on the cross section (
Figure
III

15
)
C
d
k
b
f
2
1
f
A
T
c
s
s
III

23
A
s
= total cross sectional area of tensile reinforcement
f
s
= unit stress in tensile reinforcement
Figure
III

15
Strain, Stress Distributions in Stage II
d
b
h
f
kd
c
d

kd
s
c
kd/3
jd
T=A
s
f
s
C = f
c
kd b /2
24
To simplify our expressions, we define reinforcement ratio, ρ, as the ratio of the total area of the
tensile reinforcement, A
s
, to the product of b, the width of the zone in compression, and d,
effective depth.
A
s
b
d
III

24
The reinforcement ratio
is usually in between 0.3 and 2%.
Ratios close to 1% are common
.
Replacing A
s
in
Eq.
III

23
by the definition in Eq.
III

24
:
f
s
1
2
f
c
k
III

25
Using the material moduli E
s
and E
c
, we rewrite Eq.
III

25
in terms of unit strain and the ratio
n=E
s
/E
c
:
s
n
1
2
c
k
III

26
or
s
c
k
2
n
III

27
Equati
ng the right

hand terms of Eq
s
.
III

22
and
III

27
,
1
k
k
k
2
n
III

28
We
now can
solve for
k
:
k
2
n
2
2
n
n
III

29
The relative depth to the neutral axis k is usually close to 0.3.
Equation
III

29
g
ives u
s a
convenient vehicle to calculate
the neutral axis depth in a
rectangular section
with a
single layer
of reinforcement
.
Taking moments about the centroid of the compressive force,
M
A
s
f
s
d
1
k
3
III

30
25
or
M
f
s
b
d
2
1
k
3
III

31
Th
e
s
e
expression
s
allow
us to estimate the reinforcement
unit
stress in Stage II with the caveat
that the reinforcement
unit
stress is likely to be less than that calcu
lated because we have
neglected the contribution
from all concrete in tension
.
Comparing Equations
III

30
and
III

15
,
we conclude that
, in
Stage II
,
the internal lever arm jd is equal to
d
3
k
1
d
j
III

32
and
s
s
f
A
T
III

33
with fs varying depending on the magnitude of the applied moment.
We must remember that we have assumed
the concrete
unit
stress to be linearly related to
concrete strain. This assumption needs to be examined.
We can do so replacing Eq.
III

25
in Eq.
III

31
:
M
1
2
f
c
b
k
d
2
1
k
3
III

34
This expression allows us to compute the extreme fiber
unit
stress, f
c
, for a given moment. If f
c
is
less than f’
c
/2, the error associated with the nonlinearity of concrete is usually negligible.
Curvature
is
d
k
k
1
d
k
b
E
M
c
3
1
2
2
1
c
c
III

35
It can be shown that the quantity
k
1
d
k
b
3
1
3
2
2
1
is equal to the moment of inertia of the
cracked section (the demonstration is left to the student):
2
s
2
2
1
3
12
1
steel
concrete
cr
)
kd
d
(
A
n
kd
kd
b
kd
b
I
I
I
III

36
This expression represents the moment of inertia of the transformed section shown in
Figure
III

16
.
The first term on the right

hand side of this expression is the moment o
f inertia
,
about
the
neutral axis
,
of the rectangular area of concrete above it. The secon
d term is the moment of
inertia, about the neutral axis,
of an area
equal to n times
A
s
. This moment of inertia can be
26
obtained using the parallel axis theorem and
neglecti
ng the moment of inertia
of the area nA
s
with
respect to
its own
centroidal axis.
The student should realize that the neutral axis passes through the centroid of the transformed
section.
Figure
III

16
Transformed Section
Curvature can be expressed again as:
cr
c
c
I
E
M
c
III

37
The same result can be obtained from Equations
III

31
and
III

7
.
Notice the similarities between Equations
III

20
and
III

37
. They differ in that the moment of
inertia changes from I
g
(for the uncracked section
–
Eq.
III

20
) to I
cr
(for the cracked section
–
Eq.
III

37
).
b
kd
d

kd
n A
s
Remember
:
After cracking
–
in Stage II
–
the relationship between curvature and bending
moment is:
cr
c
c
I
E
M
c
and reinforcement unit stress is linearly proportional to applied mom
ent:
jd
f
A
M
s
s
with j =
(1

k/3)
and
k
2
n
2
2
n
n
Common values:
= 1%
k = 0.3
j = 0.9
I
cr
= 0.3 I
g
27
III

5.3
Limiting Crack Width
III

5.4
Limiting
Immediate
Deflections
III

5.5
Limiting Long

Term Deflections
28
III

6
Nonlinear Response
(Stage III)
and
Strength
III

6.1
Beams with a Single Layer
of Reinforcement
Stage III ranges form yielding of the tensile
reinforcement to
the point where the
limit
ing
strain
of concrete
is reached
or strain
hardening in the reinforcement
initiates
.
In
the following discussion we ignore strain
hardening and a
ssume that the reinforcement
is elasto

plastic (Sec.
II

2.2
).
We focus on steel reinforcement although
other materials, such as fiber reinforced
plastics and glass, are used. The basic theory
would apply in all cases but all i
nferences we
make in reference to the specific shape of the
stress

strain curves assumed (
Chapter II
)
would not apply.
Figure
III

17
shows how unit strains and
stresses change as bending moment and
im
posed deformation increase. Observe that
after yielding, the stress in the reinforcement
remains constant at fy. And from step 3
through 6, the shape of the distribution of unit
stress in the concrete changes from a shape
close to a triangle to a shape b
ounded by a
parabola and an approximately straight line
(See
Chapter II
for a description o the typical
unit stress

unit strain curve assumed for
concrete).
We know that the basic equations describing
the conditions of equilib
rium (
III

14
,
III

15
)
depend on the location and magnitude of
resultants of stresses. Both location and
magnitude are functions of the shape of the
stress distribution
–
which, in Stages III and
IV, chan
ges from one loading step to the next
.
We conclude that we cannot propose
a single closed

form expression relating moment and stress
or moment and curvature
in
S
tage III
as we did for Stages I and II.
We have to look at a snapshot
of the response at a ti
me.
We therefore concentrate our attention on determining the moment at
the
point where the
beam reaches its maximum moment. We assume that, at this point, the
maximum
strain
in the
concrete
is 0.003 (Step 5 in
Figure
III

17
).
It is conventional to assume
y
3
c
Unit Strain
Unit Stress
in Steel
Unit
Stress in
Concrete
Moment
Deflection
Sta
ge I
Stage II
Stage III
M
cr
M
y
1
2
3
4
5
6
1
2
3
4
5
6
0.004
0.004
0.004
f
k f'
Figure
III

17
29
that at this strain, concrete starts
spalling,
1
although concrete has
been observed in the laboratory to
undergo larger strains before
spalling
.
W
e follow essentially the same
procedure as we did earlier but this
time in refe
rence to the conditions
in
Figure
III

19
.
The extreme

fiber
strain in concrete is set at the
useful limit of compressive strain,
ε
cu
(0.003).
(a) section
(b) unit strain
(c) concrete
(d) resultants
unit stress
Figure
III

19
Assumed Distributions of Stress and Strain at Maximum Moment
From
the
geometry
of the assumed distribution of strains
(
Figure
III

19
b
), we obtain a
relationship between the useful limit of strain and the strain in th
e reinforcement,
su
cu
1
k
u
k
u
III

38
ε
cu
= useful limit of concrete compressive strain
ε
su
= strain at centroid of reinforcement compatible with ε
cu
k
u
= ratio of depth between extreme fiber in compression and neutral axis to the effective depth
of the reinforcement
Next we consider the eq
uilibrium of
internal forces
assumed to be acting on the section. The
compressive force may be obtained by integrating the
stresses
from the neutral axis to the
1
We shall see later that if the reinforcement ratio is very small

such as 0.002

,
steel fracture may be more
critical than spalling of concrete.
d
b
h
k f
'
k d
c
d

k d
su
c
k k d
jd
T=A
s
f
y
k [k
f
'
k d b]
u
u
2
u
u
3
c
3
1
Figure
III

18
Concrete Spalling (Top Face of Beam)
30
extreme fiber in compression
but the conventional
procedure has been to do this
by assuming thr
ee
approximate coefficients
(dimensionless) to define the
properties of an “equivalent
compression block
”
commonly know as the
Whitney Stress Block (REF).
These
coefficients
are:
k
1
= Ratio of the area within
the compressive stress
distribution to an enc
losing
rectangle. This factor is
assumed to be 0.85 for concretes having strength
s
of 4,000 psi or less. For higher compressive
stresses, it is reduced linearly 0.05 for each 1000 psi. For example, it would be assumed to be
0.75 for a concrete with a comp
ressive strength of 6,000 psi.
k
2
= Ratio of the distance between the centroid of the compressive force and the extreme fiber in
compression to the neutral

axis depth. This is assumed to be k
1
/2, or 0.42 for concretes having a
strength of 4,000 psi or
less
2
.
k
3
= Ratio of the maximum compressive stress in the compression block to strength determined
from 6*12

in. test cylinders
3
. This value is set at 0.85.
Using the coefficient
s
k
1
and k
3
, we determine the resultant of compressive stresses in the
co
ncrete,
C = k
1
*k
3
*f’
c
*b*k
u
*d
III

39
This result can be visualized as the
volume of the object shown in
Figure
III

21
a. Observe that
,
given the definition of k1 shown in
Figure
III

20
,
this object has the
same volume as the box
shown in
Figure
III

21
b.
And because k
2
is assumed to be k
1
/2, the vertical distance from the top
face to the centroid is the same for both objects. We
conclude that the stress and strain conditions
at maximum moment can be idealized as shown in
Figure
III

22
.
2
The factor k
2
would be 1/3 for a linearly varying distribution of stress and ½ for a constant distribution of
stress. It may be noted that the adopted value is the mean of these two coeffic
ients.
3
The value used for this coefficient comes from an extensive experimental study, in the 1930’s, of the
strength of reinforced concrete columns. In that study, it was found that strength of concrete in the column
was less than that in the cylinder,
a difference attributed primarily to the variation in water/cement ratio in
the column caused by migration of water to the top of the vertically cast column (see Sec.
II

3.3
). There
should be no such effect in beams, but the co
nstant is used anyway.
1
1
Shape of Assumed
Stress

Strain Curve
k
1
Area of Shaded
Region
Figure
III

20
Definition of k
1
31
(a)
(b)
Figure
III

21
Equivalent Stress Bl
ock
(a) section
(b) unit strain
(c) concrete
(d) resultants
unit stress
Figure
III

22
Idealized Distributions of Stress and Strain at Maximum Moment
We should keep in mind
that, strictly speaking, our idealization does not apply to cases in which
the width of the zone in compression varies.
Assuming
that the steel yields before the maximum strain in the concrete reaches 0.003,
t
he force
in the reinforcement
, defined
using t
he reinforcement ratio
, is
:
T = ρ*fy*b*d
III

40
Equating the tensile and compressive forces acting on the section (tensile strength of concrete is
ignored in conventional analysis),
ρ*fy*b*d = k
1
*k
3
*f’
c
*
b*k
u
*d
III

41
leading to
f
y
k
1
k
3
f
c
k
u
III

42
k f
'
c
3
k d
u
k f
'
c
3
k k d
1
u
k k d
2
u
b
b
centroid
d
b
h
k f
'
k d
c
d

k d
su
c
k k d
jd
T=A
s
f
y
k [k f
'
k d b]
u
u
1
u
u
3
c
3
1
k k d
1
u
2
32
Eq
uation
III

42
gives us the
relative
depth to the neutral axis associated
with a maximum strain in
the concrete ε
cu
= 0.003. The results it yields are valid if the stress in the reinforcement is indeed
equal to f
y
.
That can be checked using Eq.
III

38
( with ε
cu
= 0.003)
. If
y
<
su
we conclude f
su
=f
y
.
If
y
<
su
,
f
y
should be replaced with
su
*E
s
in Eq.
III

41
.
If the limits defined in Section
III

6.5
are met, that is, if the beam does not have too much reinforcement, the assumption f
su
=f
y
is
c
orrect. In a beam with too much reinforcement
–
and over

reinforced beam
–
the concrete tends
to spall before the steel yields.
Taking moments about the centroid of the compressive force,
M
n
= As*fy*d*(1

k
2
*k
u
)
III

43
We can rewrite Eq.
III

43
as:
M
n
= As*fy*j*d
III

44
with j=(1

k
2
*k
u
).
F
or concrete with compressive strength between 4000 and 6000 psi, fy=60ksi,
and 0.5%<
<2%,
j varies between 0.82 and 0.97
.
T
o assume j
=
0.9 is convenient for preliminary
proportioning.
In some countries, in fact, the designer is not required to check his/her
assumptions about j. That is not the case in the US.
Equations
III

43
and
III

44
define flexural capacity. The results they produce are a direct
consequence of the criterion we have used to define capacity. This criterion sets a limit for the
maximum compressive strain. We thus
define flexural capacity by limiting strain (
Figure
III

7
).
x
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SP 05/28/08
33
III

6.2
Comparison between Stages I, II, and II
Observe the similarities between Eq. 22 and Eq. 14. Equat
ions 14 and 22 simply say, in
mathematical terms, that bending moment is resisted by an internal couple; a compressive force
and a tensile force of equal magnitude and at a distance from one another (that we shall refer to as
“internal lever arm”) directly
proportional to the “effective depth” of the beam d. Eq. 14 is
applicable in the linear range of response, and Eq. 22 is applicable at “ultimate” (when the strain
in the concrete reaches its limit). We shall use Eq. 14 to deal with problems related to s
ervice (or
unfactored) loads and Eq. 22 for problems related to factored loads.
III

6.3
Balanced Failure
III

6.4
Beams with Compression Reinforcement
III

6.5
Reinforcement Limits
III

6.6
T

Beams
III

7
Design of Elements to Resist Flexure
1)
Select Depth as L/12
2)
Select b
3)
Compute
ultimate moment
4)
Assume j=0.9
5)
Select As
6)
Compute
, check vs
min and
max
7)
Compute Mn with “actual” j
8)
Check
Mn vs Mu
III

8
Examples
34
III

8.1
Simple Beam Design
III

8.2
Deflection of a
Beam in a Frame
III

8.3
One Way Slab
III

9
Interactive Exercises
III

10
Exercises
III

10.1
Design, Construction
, and Test of a Small

Scale Beam
35
Chapter IV
Bend
ing and Axial Load: Columns
IV

1
The B
asics: Columns without Longitudinal Reinforcement
IV

2
Columns with
Longitudinal Reinforcement
IV

3
The Idea B
ehind Prestressed Concrete
IV

4
Examples
IV

5
Interactive Exercises
IV

6
Exercises
36
Chapter V
How Concrete and Steel I
nteract: Bond
V

1
Anchorage and Development
Cover
Hooks
V

2
Lap Splices
V

3
Reinforcement Layout
V

4
Examples
V

5
Interactive Exercises
V

6
V

1
Exercises
37
Chapter VI
Shear
VI

1
The
Initial
State of Stresses
and Mohr’s Circle
VI

1.1
VI

2
The Facts: Test Results
Ric
hart, ASCE database, Japanese Data
VI

3
The Myth
: The
Imaginary
Truss
VI

4
The Fable of Vc and Vs
VI

5
The Effect
of Axial Load
VI

6
Examples
VI

7
Interactive Exercises
VI

8
VI

9
Exercises
38
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