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d
t
b
V
f
f
d
t
b
v
t
w
z
V
2
=
f
f
z
v
t
f
=
I
4
b
y
I
4
y
s
f
1
2
3
t
s
w
f
t
y
w
d
f
5
4
6
2
d
t
b
d
w
V
V
y
f
f
f
v
t
w
z
z
2
=
+
I
4
I
8
y
z
(a) I-section
(b) Shear flow distribution
Figure 5.12
Shear flow distribution in an I-section.
δ
A
δ
x
0
M
y
3
(
+
δ
)δ
A
0
2
M
y
δ
(
t
f
)
23
δ
x
M
y
+
(
t
f
)
21
δ
x
y
x
2
(
t
w
)
24
δ
x
2
0
z
1
(
t
f
)
21
(
t
f
)
23
1
2
3
(
t
w
)
24
δ
x
V
2
z
(
t
w
)
24
(a) Sign of shear flow
4
4
(b) Flange-web junction
(c) Flow analogy
Figure 5.13
Horizontal equilibrium considerations for shear flow.
Afreeendofacross-sectionelementisthespecialcaseofaone-elementjunction,
for which equation 5.20 reduces to
τ
v
t
=
0
(5.21)
This condition has already been used (at
s
=
0) in deriving equation 5.16.
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