Civil Engineering Reference
In-Depth Information
Figure 2.1
Truss model in bending
A
s
f
s
. The concrete area within the compression stress block is also assumed to concentrate
at the location of the resultant
C
. This concrete element is known as the compression stringer
and is capable of resisting the resultant
C
. The distance between the tension and compression
stringers is designated
d
v
, which, of course, is equal to
jd
.
For under-reinforced members, the steel yields before the concrete crushes. At the yield
condition the steel stress
f
s
becomes
f
y
, where
f
y
is the yield strength of the tensile steel. To
simplify the analysis, the ACI Code allows the replacement of the curved concrete stress block
by a rectangular one, shown by the dotted line in Figure 2.1(b). The stress of the rectangular
stress block is 0
85
f
c
and the depth is
β
1
is taken as 0.85 for
f
c
=
27.6 MPa
(4000 psi) or less, and decreases by 0.05 for every increment of 6.9 MPa (1000 psi) beyond
f
.
β
1
c
. The coefficient
27.6 MPa (4000 psi). Based on this equivalent rectangular stress block, the magnitude of
the resultant
C
=
85
f
c
b
=
0
.
β
1
c
and the coefficient
k
2
=
β
1
/
2. Moment equilibrium shown in
Figure 2.1(b) gives:
A
s
f
y
d
1
−
β
1
c
2
d
M
=
(2.2)
where
c
is obtained from the force equilibrium,
T
=
C
:
A
s
f
y
c
=
(2.3)
0
.
85
f
c
b
β
1
Equations (2.2) and (2.3) show that the lever arm
jd
is equal to (1
−
β
1
c
/
2
d
)
d
. For normal
flexural members with steel reinforcement of 1-1.5%,
jd
or
d
v
is approximately 0
.
9
d
.
2.1.2 Equilibrium in Element Shear
2.1.2.1 Element Shear Equations
A 2-D element subjected to a shear flow
q
is shown in Figure 2.2(a). The element has a
thickness of
h
and is square, with unit length in both directions. The reinforcing steel bars are
