1
INTRODUCTION TO THE
DESIGN
OF
REINFORCED CONCRETE ELEMENTS
By
TO
ME
S
ISOP
2
TABLE OF CONTENTS
Chapter I What Reinforced Concrete is, How it Works, and How we Design it.
...........................
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 Co
nfinement
................................
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................................
................
11
II

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

4 Exercises
................................
................................
................................
.............................
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 Relat
ionship between Curvature and Bending Moment
................................
............
17
III

4 Stages of Response
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19
III

5 Linear Response (Stages I and II) and Servicea
bility
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.......................
20
III

5.1 Response before Cracking (Stage I)
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................................
...........
20
III

5.2
Response after Cracking (Stage II)
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.....
22
III

5.3 Limiting Crack Width
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................................
................................
26
III

5.4 Limiting Immediate Deflections
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................................
................
26
III

5.5 Limitin
g Long

Term Deflections
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...............
26
III

6 Nonlinear Response (Stage III) and Strength
................................
................................
....
27
III

6.1 Beams with a Single Layer of Reinforceme
nt
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...........................
27
III

6.2 Comparison between Stages I, II, and II
................................
................................
....
32
III

6.3 Balanced Failure
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................................
.........
32
III

6.4 Beams with Compression Reinforcement
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................................
..
32
III

6.5 Reinforcement Limits
................................
................................
................................
.
32
III

6.6 T

Beams
................................
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................................
.....................
32
III

7 Design of Elements to Resist Flexure
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32
III

8 Examples
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...........................
32
III

8.1
Simple Beam Design
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................................
..
33
III

8.2 Deflection of a Beam in a Frame
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...............
33
III

8.3 One Way Slab
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33
III

9 Interactive Exercises
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.........
33
3
III

10 Exercises
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................................
.........................
33
III

10.1 Design, Construction, and Test of a Small

S
cale Beam
................................
...........
33
Chapter IV Bending and Axial Load: Columns
................................
................................
............
34
IV

1 The Basics: Columns without Longitudinal Reinforcement
................................
.............
34
IV

2 Columns with Longitudinal Reinforcement
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................................
......
34
IV

3 The Idea Behind Prestressed Concrete
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34
IV

4 Examples
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...........................
34
IV

5 Interactive Exercises
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.........
34
IV

6 Exercises
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...........................
34
Chapter V How Concrete and Steel Interact: Bond
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.......
35
V

1 Anchorage and Development
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35
V

2 Lap Splices
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35
V

3 Reinforcement Layout
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35
V

4 Examples
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................................
................................
............................
35
V

5 Interactive Exerc
ises
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................................
................................
..........
35
V

6 V

1
Exercises
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..................
35
Chapter VI Shear
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...........................
36
VI

1 The Initia
l State of Stresses and Mohr’s Circle
................................
................................
36
VI

1.1
36
VI

2 The Facts: Test Results
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.....
36
VI

3 The Myth: The Imaginary Truss
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.......................
36
VI

4 The Fable of Vc and Vs
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....
36
VI

5 The Effect of Axial Load
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................................
..
36
VI

6 Examples
................................
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................................
...........................
36
VI

7 Interactive Exercises
................................
................................
................................
.........
36
VI

8
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...........
36
VI

9 Exercises
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...........................
36
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 first structures we know of in
which stone and
iron
were used together were built by the ancient Greeks
(
Figu
re
I

1
)
.
Over the
years, the stone was replaced
by
concrete and iron
by
steel.
The
reasons for the union between
s
teel 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
composite
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
–
whil
e 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
.
Figu
re
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 concrete 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 axi
s 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 concr
ete
,
think 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 a
pplied load
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 B
eam
If we 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 i
t 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 and 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 reinforceme
nt can
also
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
Bef
ore and After 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
.
You 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
Reinforced Concrete
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 Standar
d 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 t
o 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 i
s 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
linear (
Fig
ure
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 applications 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 t
o incur tensile
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.
Fig
ure
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 compre
ssive strength
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 ind
icated 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 nearl
y
constant. 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 Ritte
r, 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 ma
de 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 a
t 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 Cree
p
II

4
Exercises
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 struct
ure can be 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, int
roduce the 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

2
. Please note
that
mo
ment is
constant 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 th
at is more 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 Subje
cted to End Moments
13
III

2.1
Definition
Consider the beam
shown in figure
XXX. 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 idealized and
“actual”
beams coincide at the
locations of the hinges but not
elsewhere.
As the idealized beam deforms,
each hinge rotates. To
accommodate this rotation, fibers
above the hinge compress and fibers
below the hinge stretch.
The 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 directly proportional to the
distance t
o the neutral axis
(Figure
ZZZ)
:
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 flexural
deformation
in a segment of the beam
.
This quantity
is call
ed curvature
and,
a
s
x tends to zero
(i.e. as we
Figure
III

3
Definition of Curvature
Fiber
A
B
x/2
x/2
i
n
y
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
Curvature
Distance
Deflection
Distance
Approx. Deflection
Distance
Hinge
Rigid Link
2
1
t
B/A
B
B
Figure
III

2
Idealization of a
Deformed Beam.
14
subdivide the beam in more segments
–
Figure YYY),
curvature approaches the second derivative
of the function describing the deflected shape of the beam.
Recall from calculus that functions
with a positive
second derivative are concave, that is, shaped as a bowl (
Figure
III

4
). 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 strains, 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 include terms related to the applied
loads.
If our idealizations (Figure XXX) s
eem a bit too far from reality consider the photographs in
Figure XX. 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
mak
e us realize that we are not dealing with a continuum and
that when we compute
or measure
curvature or strain in concrete we
shall consider our results
average
values
.
Recall:
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
).
15
Figure
III

4
Second Derivative
III

2.2
The Relationship b
etween Curvature and Deflection
For the sake of illustration, c
onsider again the beam in Figure KKK. For small deformations, the
deviation of 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 A
can be obtained dividing this result by L:
4
1
4
3
2
1
0
Similarly, the deviation of B (mid

span)
with respect to
the tangent at
A is:
L
4
1
t
1
A
/
B
and, therefore, t
he deflectio
n 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)
16
L
2
2
1
B
x
f(x)
Positive Curvature
Nega
tive Curvature
16
This
expression
is, of course, a course approximation. We shall evaluate it numerically
but
simply
for illustration purposes.
Nevertheless, this expression d
oes
demonstrate that
there is a
unique relationship betwee
n curvature distribution and deflected shape
. If one is given, the other
can be computed.
That is what is of importance.
If we use more hinges:
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
Recall:
T
here is a unique relationship between curvature distribution and deflected shape
.
To compute deflections we, therefore, need to estimate the distribution of
curvature.
The c
hange in slop
e
between points A and B
is equal to the area under
the
curvature
diagram
between A and B
.
T
he deviation of
point
B with respect to the tangent at
point
A
is equal to the
mom
ent
with respect to B
of the area between A and B.
17
III

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

2
)
.
In
design
,
it is important
to relate strains to
bending moment
because we judge whether an element is approaching f
ailure by comparing
computed strains with strains observed to be associated with failure in uniaxial 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

5
)
.
Figure
III

5
Curvature, Strain, Moment
To understand how a beam resists bending
moment and how the mechanism of resistance
is related to cu
rvature, consider first the craw
bar shown in
Figure
III

12
. 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

8
Equilibrium of moments requires:
d
j
T
x
F
M
III

9
Strain
Moment
Curvature
Sec.
III

3
Sec.
III

2
Design
(Sec.
III

6
)
x
C
=
T
T
F
F
j
.
d
F
Figure
III

6
Craw Bar
18
Consider
now
the beam in
Figure
III

13
.
Consider the
equilibrium of the segment
to
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 that b
elow the neutral axis
there are
tensile
unit
stress
es
. Above
it
,
there mus
t 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
.
Again
, equilibrium of
forces and
moment
s
require
:
T
C
III

10
a
nd
d
j
T
M
III

11
The
magnitude of the resultants C and T
that balance
the moment M
and the distance between them (jd)
depend on the
dist
ribution 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

11
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

14
.
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 causes 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, curvature,
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
.
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
Figure
III

7
Internal Forces Associated
with
Bendi
ng
C
C
Figure
III

8
19
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

2
, it may go
through four different stages of
response.
Initially (M<M
cr
in
Error! Reference source not
found.
), the element responds linearly (an increase
in moment is directly proportional to an increase
in displacement). 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 virtually no change in moment. Depending on the amount of te
nsile 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
Error! Reference source not found.
.
OVER

REINFORCED BEAMS?
Figure
III

10
Typical Moment

Deflection Relationship
T
he relationship between moment
, curvature,
and strain
is different in 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 interested in Stage IV, the range where the moment may start
increasing a
gain. 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
Cracking
Moment
Deflection
Figure
III

9
Cross Section of Beam
with Single Layer of Reinforcement
d
b
Tensile
Reinforcement
Area = A
s
h
20
III

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

3
we discussed that effective design requires consideration of the response of the
reinforced concrete elemen
t 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
and strains
they
produce
because these relationships are useful in the d
evelopment 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

15
)
is uncracked and that the effe
ct 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 over depth of section is linear (
Figure
III

17
b
).
2.
The normal
unit
stress is linearly related to
unit
strain in tension and compression
(
Figure
III

17
c
)
by
the
Young’s Modulus
for Concrete, E
c

obtained from 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) will be, according to our idealization, at mid

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

10
and
III

11
apply in all the stages of response, t
he internal forces and resisting moment can be expressed i
n
te
rms of the extreme

fiber
unit
stress
f as
:
c
c
f
h
b
4
1
2
h
b
f
2
1
T
C
III

12
c
2
f
h
b
6
1
h
3
2
T
M
Figure
III

11
21
III

13
The factor 2/3h in Equation
III

13
is the distan
ce between the resultants of compressive and
tensile stresses
; the internal lever arm jd:
h
3
2
d
j
III

14
This
result
is a
consequence
of the assumed distribution of
unit
stress.
Unit stre
ss 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

15
And c
urvature
is
g
c
c
I
E
M
c
III

16
where
3
12
1
g
bh
I
and c = h/2
.
From mechanics of materials, we know the
relationship
described by Eq.
III

16
to be true.
The boundary between Stages I and II is given by the moment at crackin
g 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

17
Here f
r
is modulus of rupture (Sec.
II

3.2
).
Recall:
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
mome
nt does not exceed:
c
2
r
2
cr
'
f
4
5
h
b
f
h
b
6
1
M
22
III

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

17
).
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

18
, 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

18
), we obtain the relationship
s
c
1
k
k
III

18
s
= unit strain in reinforcement
c
= unit strain in concrete at the extreme fiber in compression
k = ratio of neutral

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

18
)
C
d
k
b
f
2
1
f
A
T
c
s
s
III

19
A
s
= total cross se
ctional area of tensile reinforcement
f
s
= unit stress in tensile reinforcement
Figure
III

12
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
23
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 c
ompression, and d,
effective depth.
A
s
b
d
III

20
Replacing A
s
in
Eq.
III

19
by the definition in Eq.
III

20
:
f
s
1
2
f
c
k
III

21
Using the material moduli E
s
and E
c
, we rewrite Eq.
III

21
in terms of unit strain and the ratio
n=E
s
/E
c
:
s
n
1
2
c
k
III

22
or
s
c
k
2
n
III

23
Equati
ng the right

hand terms of Eq
s
.
III

18
and
III

23
,
1
k
k
k
2
n
III

24
We
now can
solve for
k
:
k
2
n
2
2
n
n
III

25
Equation
III

25
g
ives us a convenient vehicle to calculate
the neutral axis depth in a
rectangular
sec
tion
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

26
or
24
M
f
s
b
d
2
1
k
3
III

27
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 calculated because we have
neglected the contribution
from all concrete in tension
.
Comparing Equations
III

26
and
III

11
,
we conclude that
, in
Stage II
,
the internal lever arm jd is equal to
d
3
k
1
d
j
III

28
and
s
s
f
A
T
III

29
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 assum
ption needs to be examined.
We can do so replacing Eq.
III

21
in Eq.
III

27
:
M
1
2
f
c
b
k
d
2
1
k
3
III

30
This expression allows us to compute the extre
me 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

31
It can b
e 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
3
cr
)
kd
d
(
d
b
n
kd
b
3
1
I
III

32
This expression repres
ents the moment of inertia of the transformed section shown in
Figure
III

13
.
The first term on the right

hand side of this expression is the moment of 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
obtained using the parallel axis theorem and neglecti
ng the moment of inertia
of the area nA
s
with
respect to
its own
centro
idal axis.
25
The student should realize that the neutral axis passes through the centroid of the transformed
section.
Figure
III

13
Transformed Section
Curvature can be expressed
again as:
cr
c
c
I
E
M
c
III

33
The same result can be obtained from Equations
III

27
and
III

7
.
Notice the similarities between Equations
III

16
and
III

33
. They differ in that the moment of
inertia changes from I
g
(for the uncracked section
–
Eq.
III

16
) to I
cr
(for the cracked section
–
Eq.
III

33
).
b
kd
d

kd
n A
s
Recall:
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 moment:
jd
f
A
M
s
s
with j =
(1

k/3)
and
k
2
n
2
2
n
n
26
III

5.3
Limiting Crack Width
III

5.4
Limiting
Immediate
Deflections
III

5.5
Limiting Long

Term Deflections
27
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
reinforceme
nt 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 assume 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 inferences we
make in reference to the specific shape of the
stress

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

20
shows how unit strains and
stresses change as bending moment and
imposed deformation increase. Observe that
after yielding, the stress in the re
inforcement
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 bounded 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 equilibrium (
III

10
,
III

11
)
depe
nd 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, changes from one loading step to the next
.
We conclude that we cannot propose
a s
ingle 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 time.
We therefore concentrate our attention on determining the moment at
the
p
oint 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

20
). It is conventional to assume
y
3
c
Unit Strain
Unit Stress
in Steel
Unit
Stress in
Concrete
Moment
Deflection
Stage 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

14
28
that at this strain, concrete starts
spalling,
1
a
lthough 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 reference to the conditions
in
Figure
III

16
.
The extre
me

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

16
Assumed Distributions of Stress and Strain at Maximum Moment
From
the
geometry
of the assumed distribution of strains
(
Figure
III

16
b
), we obtain a
relationship between the useful limit of strain and the strain in the reinfor
cement,
su
cu
1
k
u
k
u
III

34
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

15
Concrete Spalling (Top Face of Beam)
29
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

35
This result can be visualized as the
volume of the object shown in
Figure
III

18
a. Observe that
,
given the definition of k1 shown in
Figure
III

23
,
this object has the
same volume as the box
shown in
Figure
III

18
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

19
.
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

17
Definition of k
1
30
(a)
(b)
Figure
III

18
Equivalent Stress Bl
ock
(a) section
(b) unit strain
(c) concrete
(d) resultants
unit stress
Figure
III

19
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

36
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

37
leading to
f
y
k
1
k
3
f
c
k
u
III

38
k f
'
c
3
k d
u
k f
'
c
3
k k d
1
u
k k d
2
u
b
b
ce
ntroid
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
31
Eq
uation
III

38
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

34
( 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

37
.
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

39
We can rewrite Eq.
III

39
as:
M
n
= As*fy*j*d
III

40
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

39
and
III

40
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

5
).
x
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SP 05/28/08
32
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
33
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
34
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
35
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
36
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
37
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