Civil Engineering Reference
In-Depth Information
3
bkd
(
)
1000
×
44.41
(
)
2
(
)
2
I
=
+
nA
d d
−
=
+××
8.51
887
175
−
44.41
cr
s
3
3
8
4
=
157923864.61.579
=
×
10
mm
(
)
6
18.72
×
10
−
××
=
200
44.41
ε=
0.000785
bi
8
4700
25
4.110
0.7
0.7
0.95
0.0167
0.000785
0.01189
ε= ε+ε= ×
C
×
+
=
fd
E u
bi
Applying the linear regression equation:
M
M
M
M
60.11
53.22
u
n
== =
1.1295
u
n
M
M
ρ
ρ
f
f
d
d
n
ff
f
=
0.7815
+=
11.1295
n
sy
λ=
ρ×
ρ
0.7
f
d
d
0.1295
×
0.00507
×
400
×
175
f
fu
f
ρ=
=
0.000214
f
f
0.7815
×× ×
0.70.95
2068
×
200
sy
2
A d
=ρ =
37.41 mm /m
f
f
Each NSMtapehas theareaof32mm.
2
No. of NSM tapes/m
=
37.41
32
=
1.17 tapes
Spacing
=
1000
1.17
=
855 mm
Use one NSM tape @ 850 mm.
No. of tapes per meter
=
1000
850
=
1.176 tapes
2
2
A
f
=
32 mm
×
1.176
=
37.63 mm /m
Using force equilibrium to find the neutral axis depth (
c
):
α= +×
fbcAf
A
0.7
f
c
s y
f
fu
2
c
dc
c
dc
2
2
2
α=
ε
ε
ε
ε
ε×
−×ε
ε×
−ε
cf
cf
fd
fd
−
=
−
(
)
2
(
)
3
2
3
c
c
f
c
f
c
Into forceequilibrium:
2
2
2
c
dc
ε×
−×ε
c
dc
ε×
−ε
=+×
fd
fd
−
fbcAf
A
0.7
f
(
)
c
s y
f
fu
(
)
2
3
f
c
f
c
(
−ε
2
2
Multiply this force equilibrium equation by
dc
f
c
−
ε
2
(
)
(
)
(
)
2
2
fd
3
2
εε −
dcfbc
f bc
=
Af
+
Af
×
0.7
dc
−
ε
fd
cf
c
c
s y
f
fu
f
c
3
25
4700
1.71
0.00182
ε= ×
=
c
25
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