Civil Engineering Reference
In-Depth Information
φ ,
Dividing Equation (5.12) by
fbd
c
f
M
fbd
f
f
d
d
a
f
f
d
d
a
u
y
fe
=ρ −+Ψρ
d
d
−
(5.13)
s
f
f
f
2
2
2
φ
2
2
c
f
cf
c
f
Substituting Equation (5.11) into (5.13) and rearranging the terms,
2
a
d
−−
−Ψ
Ψ
1.176
1
a
d
2.353
f
2
Q
+
Ψ
Q
=
0
(5.14)
1
2
f
f
f
f
where
f
f
d
d
y
Q
=ρ
(5.15)
1
s
cf
M
fbd
d
d
u
Q
=
φ
+Ψ−
Q
(5.16)
2
1
f
2
c
f
f
When
Ψ=
is substituted into Equation (5.14),
0.85
f
2
a
d
a
d
[
]
−−
2
0.208
Q
+
2.77
Q
=
0
(5.17)
1
2
f
f
or
a
d
1
2
(
)
(
)
2
=
2
− −− −
0.208
Q
2
0.208
Q
11.08
Q
(5.18)
1
1
2
f
Now, to establish a range of values for
Q
1
,
f
=
3000
−
8000 psi
c
f
=−
40
75 ksi
y
33, 000
75, 000
200
75, 000
−
3
−
3
ρ= =×<
2.19
10
=
2.67
×
10
controls
s
,min
or
38, 000
40, 000
200
40, 000
−
3
−
3
−
3
ρ= =×>
6.71
10
=
5.0
×
10
,
6.71
×
10
controls
s
,min
f
f
87
87
c
ρ=ρ= ×β
+
0.75
0.75
0.85
for
ε≈
0.004 (ductile failure)
s
,max
b
1
st
f
y
y
8, 000
40, 000
87
= ×××
+
0.75
0.85
0.65
=
0.0568
87
40
Search WWH ::
Custom Search