Civil Engineering Reference
In-Depth Information
Figure 3.8
Moment and curvature diagrams for simply supported beams under uniform load
Equation (3.50) satisfies the two limiting cases. First,
I
=
I
g
when
M
cr
/
M
=
1. Second,
I
M
approaches zero. The moment of inertia
I
obviously has a value
between
I
g
and
I
cr
. The power of 4 for (
M
cr
/
=
I
cr
when
M
cr
/
M
) in Equation (3.50) was selected to best fit
the experimental curve.
When a beam is subjected to lateral loads, the moment is no longer uniform along the length
of the beam. Figure 3.8(a) shows a simply supported beam under uniform load. According
to the parabolic bending moment diagram (Figure 3.8b), the moment in the central region of
the beam exceeds the cracking moment
M
cr
, causing cracks to develop and the curvature
to
increase rapidly (Figure 3.8c). Near the ends, however, the regions are still uncracked since
the moment is still less than the cracking moment. For such a beam with both cracked and
uncracked regions, a rigorous calculation of the deflection using numerical integration would
be very tedious. For practical purposes, an effective moment of inertia
I
e
was proposed by
Branson (1977) and adopted by the ACI Code, Section 9.5.2.3 (ACI 318-08):
φ
1
3
I
cr
M
cr
M
a
3
M
cr
M
a
I
e
=
I
g
+
−
≤
I
g
(3.51)
where
M
a
is the maximum moment at midspan, and the cracking moment is
f
r
I
g
y
t
M
cr
=
(3.52)
In Equation (3.52)
y
t
is the distance from the centroidal axis of gross section to the extreme
fiber in tension.
f
r
is the modulus of rupture defined for normal-weight concrete by
63
f
c
(
MPa
)or
f
r
=
5
f
c
(
psi
)
f
r
=
0
.
7
.
(3.53)
