Civil Engineering Reference
In-Depth Information
be calculated. A suitable beam section may then be chosen from handbooks which list
properties and dimensions, including section moduli, of standard structural shapes.
The selection of a beam cross section depends upon many factors; these include the
type of loading and construction, thematerial of the beamand several others. However,
for a beam subjected to bending and fabricated frommaterial that has the same failure
stress in compression as in tension, it is logical to choose a doubly symmetrical beam
section having its centroid (and therefore its neutral axis) at mid-depth. Also it can be
seen from Fig. 9.5(b) that the greatest values of direct stress occur at points furthest
from the neutral axis so that the most efficient section is one in which most of the
material is located as far as possible from the neutral axis. Such a section is the I-section
shown in Fig. 9.2.
E
XAMPLE
9.1
A simply supported beam, 6m long, is required to carry a uniformly
distributed load of 10 kN/m. If the allowable direct stress in tension and compression
is 155N/mm
2
, select a suitable cross section for the beam.
From Fig. 3.15(d) we see that the maximum bending moment in a simply supported
beam of length
L
carrying a uniformly distributed load of intensity
w
is given by
wL
2
8
M
max
=
(i)
Therefore in this case
6
2
×
10
M
max
=
=
45 kNm
8
The required section modulus of the beam is now obtained using Eq. (9.13), thus
10
6
155
M
max
σ
x
,max
=
45
×
290 323mm
3
Z
e,min
=
=
From tables of structural steel sections it can be seen that a Universal Beam, 254mm
×
102mm
28 kg/m, has a sectionmodulus (about a centroidal axis parallel to its flanges)
of 307 600mm
3
. This is the smallest beam section having a section modulus greater
than that required and allows a margin for the increased load due to the self-weight of
the beam. However, we must now check that the allowable stress is not exceeded due
to self-weight. The total load intensity produced by the applied load and self-weight is
×
28
×
9
.
81
10
3
10
+
=
10
.
3kN
/
m
Hence, from Eq. (i)
6
2
10
.
3
×
M
max
=
=
46
.
4kNm
8
Therefore from Eq. (9.13)
10
3
10
3
46
.
4
×
×
150
.
8N
/
mm
2
σ
x
,max
=
=
307 600
