# INTRODUCTION TO THE DESIGN OF REINFORCED CONCRETE

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1

INTRODUCTION TO THE
DESIGN
OF
REINFORCED CONCRETE

By
TO
ME
S

ISOP

2

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

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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

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11

II
-
4 Exercises

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11

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

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11

II
-
4.2 Axial Load vs. Deformation for a Column with Reinforcement

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.

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

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15

III
-
3 The Relationship b
etween Curvature and Bending Moment

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18

III
-
4 Stages of Response

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20

III
-
5 Linear Response (Stages I and II) and Serviceability

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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

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....

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

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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

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...........

34

Chapter IV Bending and Axial Load: Columns

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............

35

IV
-
1 The Basics: Columns without Longitudinal Reinforcement

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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

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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

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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

midspan

. If the beam is not reinforced, it fails when the applied lo
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

(dashed line)

of the bar
s

does not

crack prevention
.
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
and
to
reach deflections
several

times larger
than the deflection at first cracking (Point B)
.

7

Figure
I
-
4

-
Deflection Relationship for a Beam with and without Reinforcement

The implications of the
phenomena

described
are tremendous.
on of reinforcement,
the beam i
s weak a
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

of reinforcement have limits. Too much or too little reinforcement

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
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

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
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

Ritter defined the initial portion of the stress
-
strain curve for concrete by a parabola known as the
Ritter Parabola

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)

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
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

Figure
III
-
1

Beam Subjected to End Moments

13

III
-
2.1
Defin
ition

Consider the beam
shown in
Figure
III
-
2
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

Distance

Deflection

Distance

Approx. Deflection

Distance

Hinge

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!
).

Functions with a negative second derivative are convex.

If we use
the
dist
ance

to the extreme fiber in compression

d
y
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

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
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

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

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 related to daily use of the structure

should remain within the linear ranges of
response for concrete
(in compression) and steel. Service
-
level
unit
stress
es or strains approaching those associated with failure of the materials.
It is important
that
we understand the relationships between service
-
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
,

the
neutral axis
,

of the rectangular area of concrete above it. The secon
d term is the moment of
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
.
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
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

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

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

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

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

VI
-
6

Examples

VI
-
7

Interactive Exercises

VI
-
8

VI
-
9

Exercises

38