Civil Engineering Reference
In-Depth Information
Solution:
The unstrengthened beams have no steel stirrups provided in the clear span.
28
2
356 mm
d
=+−−−=
100
305
25
10
28
2
305
25
10
d
fv
=−−− =
256 mm
f
35
6
c
V
=
bd
= ××=
150
356
52,653.1N
= <=
52.65 kN
V
90 kN
O.K.
c
w
exp
6
Afd
S
fv fe
fv
Ψ= ×
V
0.85
ff
f
3,790
228,000
*
ε=ε= ×
C
0.95
=
0.0158
CFRP interior exposure
fu
E u
2/ 3
35
27
k
=
=
1.19
1
23,300
10.165
L
e
=
=
51.71mm
(
)
0.58
3
×
×
228
×
10
256
51.71
256
−
k
=
=
0.8
2
1.19
××
×
0.8
51.71
k
v
=
=
0.262
<
0.75
11,900
0.0158
ε=ε= ×
k
0.262
0.0158
=
0.00414
>
0.004
fe
v u
20.165
×
×
50
×
228,000
×
0.004
×
256
f
Ψ= ×
0.85
=
26,195.6N
=
26.2kN
125
VVVV
=++ψ =
52.65 kN
+ +
0
26.2kN
= <=
78.85 kN
V
162 kN
O.K.
n
c
s
f
f
exp
The ACI 440 model significantly underestimates the shear capacity with FRP in
this case, which is considerably on the conservative side. However, it is important
to note here that Beam BT5, which is identical to Beam BT4 except for using side
strips, failed at a much lower capacity (121.5 kN). It is worthwhile to check the
capacity of BT5 using the ACI440.2R-08 model (Problem 6.5 below) to see if it
captures a similar drop in shear strength. The reader is referred to a relatively more
recent analytical model using the truss analogy method for a more in-depth com-
parison with existing models (Colotti, Spadea, and Swamy 2004). The reader is also
referred to a recent article assessing various design models for shear strengthening
(Pellegrino and Vasic 2013).
Chapter Problems
Problem 6.1
The beam section in Problem 2.5 is deficient in shear, so strengthen the beam to
resist enough shear such that it fails in flexure and not shear. The beam is under its
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