Civil Engineering Reference
In-Depth Information
(
y
,
z
) in the section. Thus
M
y
I
y
M
z
I
z
σ
x
=−
y
−
z
(ii)
Now
100 cos 30
◦
=
M
z
=
86
.
6kNm
(iii)
100 sin 30
◦
=
M
y
=
50
.
0kNm
M
z
is, in this case, a negative bending moment producing tension in the upper half
of the beam where
y
is positive. Also
M
y
produces tension in the left-hand half of
the beam where
z
is positive; we shall therefore call
M
y
a negative bending moment.
Substituting the values of
M
z
and
M
y
from Eq. (iii) but with the appropriate sign in
Eq. (ii) together with the values of
I
z
and
I
y
from Exs 9.2 and 9.3 we obtain
10
6
10
6
86
.
6
×
50
.
0
×
σ
x
=
10
6
y
+
10
6
z
(iv)
193
.
7
×
27
.
0
×
or
σ
x
=
0
.
45
y
+
1
.
85
z
(v)
Equation (v) gives the value of direct stress at any point in the cross section of the
beam and may also be used to determine the distribution over any desired portion.
Thus on the upper edge of the top flange
y
100mm, so
that the direct stress varies linearly with
z
. At the top left-hand corner of the top flange
=+
150mm, 100mm
≥
z
≥−
252
.
5N
/
mm
2
(tension)
σ
x
=
0
.
45
×
(
+
150)
+
1
.
85
×
(
+
100)
=+
At the top right-hand corner
117
.
5N
/
mm
2
(compression)
σ
x
=
0
.
45
×
(
+
150)
+
1
.
85
×
(
−
100)
=−
The distributions of direct stress over the outer edge of each flange and along the
vertical axis of symmetry are shown in Fig. 9.7. Note that the neutral axis of the beam
section does not in this case coincide with either the
z
or
y
axes, although it still passes
through the centroid of the section. Its inclination,
α
, to the
z
axis, say, can be found
by setting
σ
x
=
0 in Eq. (v). Thus
0
=
0
.
45
y
+
1
.
85
z
or
y
z
=
1
.
85
0
.
45
=
−
4
.
11
=
tan
α
which gives
76
.
3
◦
α
=
Note that
α
may be found in general terms fromEq. (ii) by again setting
σ
x
=
0
.
Hence
M
y
I
z
M
z
I
y
=
y
z
=−
tan
α
(9.14)
