Civil Engineering Reference
In-Depth Information
8
m
h
s
=
16
=
0.5 m
Application of Equations (14.26) and (14.25) gives
3(2)
2
α
d
=
0.5
+
100(0.4)
−
80(0.5)
=
0.60
1200×10
−6
2000×10
−6
l
h
0.6
f
= 16
=
11.8
Thus, the minimum thickness for beam with FRP is:
8
11.8
h
f
=
=
0.68 m;
take
h
f
=
0.7 m;
d
=
0.6 m
The bending moment at mid-span in service,
(16 kN/m)
(8 m)
2
8
M
=
=
128.0 kN.m
The permissible stress in FRP in service is
σ
f
=
E
f
ε
f
=
40 GPa (2000 ×
10
−6
)
80 MPa. Substitution of this value in Equation (14.1) with an
estimated value
y
CT
=
=
0.9d
=
0.54 m gives:
m
80 MPa (0.54)
128 kN
−
2.96×10
−3
m
2
2960 mm
2
A
f
=
=
=
0.583 m. Substitution of this value in Equation (14.1) gives a more
accurate value
A
f
Application of Equations (14.2) and (14.5) gives
c
=
51 mm and
y
CT
=
2740 mm
2
.
Example 14.2 Verification of the ratio of span to deflection
Compare
l
/
D
centre
for the beam designed in Example 14.1 with that of a
conjugate steel-reinforced beam carrying the same load intensity and
having the same thickness
h
s
=
h
f
=
0.7 m and
d
=
0.6 m, but (
l
/
h)
s
=
16.
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