Civil Engineering Reference
In-Depth Information
Plastic neutral axis
due to bending
y
s
Y
s
Y
s
Y
t
w
a
a
a
a
G
z
s
Y
s
Y
s
Y
Stress due to
axial load
Plastic neutral axis
due to bending
and axial load
F
IGURE
18.15
Combined bending
and axial
compression
(d)
(a)
(b)
(c)
additional stresses which cannot, obviously, be greater than
σ
Y
. Thus the region of
the beam section supporting compressive stresses is increased in area while the region
subjected to tensile stresses is decreased in area. Clearly some of the compressive
stresses are due to bending and some due to axial load so that the modified stress
distribution is as shown in Fig. 18.15(c).
Since the beam section is doubly symmetrical it is reasonable to assume that the area
supporting the compressive stress due to bending is equal to the area supporting
the tensile stress due to bending, both areas being symmetrically arranged about the
original plastic neutral axis. Thus from Fig. 18.15(d) the reduced plastic moment,
M
P,R
, is given by
M
P,R
=
σ
Y
(
Z
P
−
Z
a
)
(18.20)
where
Z
a
is the plastic section modulus for the area on which the axial load is assumed
to act. From Eq. (18.6)
a
2
+
2
at
w
2
a
2
a
2
t
w
Z
a
=
=
also
P
=
2
at
w
σ
Y
so that
P
2
t
w
σ
Y
Substituting for
Z
a
, in Eq. (18.20) and then for
a
, we obtain
a
=
Z
P
−
P
2
4
t
w
σ
Y
M
P,R
=
σ
Y
(18.21)
