Civil Engineering Reference
In-Depth Information
Compare to 247.6 k-ft from experiment.
Why the difference? Contribution of compression steel and steel strain harden-
ing in tension as well as
Ψ
f
=
0.85.
Unstrengthened beam capacity (
M
n
):
1.76 83.5
0.8550.8
×
×× ×
c
=
=
4.32 in.
10
c
d
4.32
16.25
=
=
0.266
<
0.375
tensioncontrolledfailure
φ =
0.9
t
−
β
c
0.84.32
2
×
1
M
=
Af
d
=
1.76
×
83.5
×
16.25
−
=
2134.15 k-in.
n
s y
2
=
177.85 k-ft
Compare to 180.4 k-ft from the control beam in experiment.
Why the small difference? Contribution of compression steel and steel strain
hardening in tension.
M
M
221.3
177.85
n
n
S.R.
== =
1.244
From the experiments:
247.6
180.4
S.R.
=
=
1.373
Example 5.4: Design
For the beam given in Example 5.3, determine the CFRP area needed to increase
the total section moment capacity by 25%.
Solution:
Unstrengthened beam capacity:
M
=× =
0.9
2134.15 k-in
1920.74 k-in
u
M
M
1.25
2400.92 k-in
=× =
u
u
Assume
dh
==
18" CFRP sheets bonded onlytosoffit).
f
A
bd
1.76
s
ρ= =
=
0.01083
s
10
×
16.25
f
f
d
d
83.5
5
16.25
18
y
Q
=ρ
=
0.01083
×
×
=
0.1633
1
s
cf
M
fbd
d
d
2400.92
0.9510
16.25
18
u
cf
Q
=
φ
+Ψ−=
×× ×
Q
+
0.1633 0.85
−
=
0.156
2
1
f
2
2
18
f
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