Civil Engineering Reference
In-Depth Information
Compare to 247.6 k-ft from experiment.
Why is there a difference? The actual
Ψ
f
=
1.0, and the small strain hardening
in tension steel is ignored here. Considering
Ψ
f
=
1.0,
0.85.54
2
×
0.85.54
2
×
M
n
=
1.76
×
83.5
×
16.25
−
+× ×
1.00.286
193.38
×
16.36
−
0.85.54
2
×
+× ×
0.22
62.45
−
1.563
=
2853.67 k-in
=
237.81 k-ft
The small difference that still exists is due to some strain hardening in the tension
steel (Rasheed et al. 2010).
Example 5.7: Design
Chaallal, Nollet, and Perraton (1998) presented a CFRP flexural strengthening
design example of a doubly reinforced simple beam (Figure 5.9). They designed
the beam using an iterative approach. The reinforced concrete section details and
material properties are shown below.
Solution:
Assume ductile crushing failure mode
Assume yielding of compression steel at failure
Assume
d
f
=
h
=
600 mm
f
f
d
d
2400
350
400
30
535
600
y
Q
=ρ
=
×
×
=
0.1524
1
s
×
535
cf
f
f
d
d
400
350
400
30
535
600
s
0.0254
Q
=ρ
=
×
×
=
1
s
×
535
cf
M
fbd
d
d
d
d
u
cf
Q
=
φ
+Ψ−−Ψ−
Q
Q
2
1
f
1
f
2
f
f
6
×
×× ×
500
10
535
600
58
600
=
+
0.1524 0.85
−
−
0.0254 0.85
−
=
0.1215
2
0.930
350
600
Concrete Properties
Steel Properties
CFRP Properties
f
c
= 30 MPa
f
y
= 400 MPa
f
fu
= 2400 MPa
E
c
= 24647 MPa
E
s
= 200 GPa
E
f
= 150 GPa
A
s
= 2400 mm
2
ε
cu
= 0.003
β
1
= 0.833
ε
bi
= Zero
t
f
= 0.17 mm
A
s
= 400 mm
2
M
u
=380 KN.m
M
u
= 500 KN.m
A
f
350 mm
FIGURE 5.9
Design of a doubly reinforced beam example by Chaallal, Nollet, and Perraton
(1998).
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