Civil Engineering Reference
In-Depth Information
Equivalent timber
reinforcing
'plates'
Equivalent steel
'joist'
d
b
t
t
E
s
E
t
E
s
E
t
E
t
E
s
t
t
b
F
IGURE
12.2
Equivalent
beam sections
(a)
(b)
From Eqs (12.1) and (12.3)
E
t
I
t
E
t
I
t
+
M
t
=
M
E
s
I
s
or
M
M
t
=
(12.6)
E
s
I
s
E
t
I
t
+
1
Substituting in Eq. (12.5) from Eq. (12.6) we have
My
I
t
+
σ
t
=−
(12.7)
E
s
E
t
I
s
Equation (12.7) could in fact have been deduced directly from Eq. (9.9) since
I
t
+
(
E
s
/
E
t
)
I
s
is the second moment of area of the equivalent timber beam of
Fig. 12.2(a). Similarly, by considering the equivalent steel beam of Fig. 12.2(b), we
obtain the direct stress distribution in the steel, i.e.
My
I
s
+
σ
s
=−
(12.8)
E
t
E
s
I
t
E
XAMPLE
12.1
A beam is formed by connecting two timber joists each
×
×
100mm
300mm placed symmetrically between
them (Fig. 12.3). If the beam is subjected to a bending moment of 50 kNm, determine
the maximum stresses in the steel and in the timber. The ratio of Young's modulus for
steel to that of timber is 12 : 1.
400mm with a steel plate 12mm
The second moments of area of the timber and steel about the centroidal axis,
Gz
, are
400
3
12
10
6
mm
4
I
t
=
×
×
=
×
2
100
1067
