Civil Engineering Reference
In-Depth Information
3.2.3.2 Design of Ductile Sections
Design problems to find areas of tension and compression steel are posed as follows:
Given:
b
,
d
,
M
u
,
f
y
,
f
c
,
ε
u
Find:
A
s
,
A
s
,
f
s
,
ε
s
and
a
(or
c
=
a
/β
1
)
The problem shows five unknowns, but we have only four equations. Therefore, one of the
five unknowns must be assumed during the design process. It will be shown below that the
most convenient unknown to choose is the depth
a
.
In the design process it is convenient to take advantage of the two-internal-couples concept
as shown in Figure 3.12(d) and (e). The bending resistance of a doubly reinforced beam
can be considered to consist of two internal couples. One couple
M
1
is contributed by the
concrete compression stress block 0
85
f
c
ba
and the other
M
2
by the compression steel area
A
s
.
The tensile steel area
A
s
is separated into
A
s
1
and
A
s
2
for
M
1
and
M
2
, respectively. Because
the first couple is identical to a singly reinforced beam which can be easily solved by the
equilibrium condition alone, the separation of the two internal couples is very convenient for
the design of doubly reinforced beams.
In the separation process we have expanded the two variables,
A
s
and
M
u
, into four variables,
A
s
1
,
A
s
2
,
M
1
and
M
2
, resulting in the addition of two unknowns. At the same time, two
additional equations stating these separations are created. These two additional equations are:
.
Separation of
M
u
M
u
=
M
1
+
M
2
(3.106)
Separation of
A
s
A
s
=
A
s
1
+
A
s
2
(3.107)
The two groups of equations and their unknown variables are summarized as follows:
First Internal Couple M
1
Type of equation
Equations
Unknowns
85
f
c
ba
Equilibrium of forces
A
s
1
f
y
=
0
.
A
s
1
a
(3
.
108)
85
f
c
ba
d
a
2
Equilibrium of moments
M
1
=
ϕ
0
.
−
M
1
a
(3
.
109)
Second Internal Couple M
2
Type of equation
Equations
Unknowns
A
s
f
s
A
s
f
s
Equilibrium of forces
A
s
2
f
y
=
A
s
2
(3
.
110)
d
)
Equilibrium of moments
M
2
=
ϕ
A
s
2
f
y
(
d
−
M
2
A
s
2
(3
.
111)
Compatibility of compression
steel
ε
s
ε
u
=
d
c
−
ε
s
(
c
=
a
/β
1
)
a
(3
.
112)
c
Constitutive law of compression
steel
f
s
E
s
ε
s
ε
s
≤
ε
y
f
s
,ε
s
=
for
(3
.
113a)
f
s
ε
s
>ε
y
f
s
,ε
s
=
f
y
for
(3
.
113b)
