Civil Engineering Reference
In-Depth Information
5.12.8 Example 8 - shear centre of a channel section
Problem
. Determinethepositionoftheshearcentreofthechannelsectionshown
in Figure 5.14.
Solution
. The shear flow distribution in the channel section was analysed in
Section 5.12.7 and is shown in Figure 5.14a. The resultant flange shear forces
shown in Figure 5.14b are obtained from
b
V
f
=
(τ
v
t
f
)
d
s
f
0
=
(
V
z
/
I
y
)
[
d
f
t
f
s
f
/
4
]
0
=
(
V
z
/
I
y
)(
d
f
t
f
b
2
/
4
)
,
while the resultant web shear force is
V
w
=
V
z
(see Section 5.12.7).
These resultant shear forces are statically equivalent to a shear force
V
z
acting
through the point S which is a distance
d
f
t
f
b
2
4
I
y
a
=
V
f
d
f
V
w
=
from the web, and so
d
f
t
f
b
2
4
I
y
b
2
t
f
2
bt
f
+
d
f
t
w
.
y
0
=
+
5.12.9 Example 9 - shear stresses in a pi-section
Problem
. The pi shaped section shown in Figure 5.42b has a shear force of 100
kN acting parallel to the
y
axis. Determine the shear stress distribution.
Solution.
Using equation 5.23, the shear flow in the segment 12 is
s
1
(τ
h
×
12
)
12
=
−
100
×
10
3
(
−
100
+
s
1
)
×
12
×
d
s
1
,
1400
×
10
4
0
whence
(τ
h
)
12
=
0.007143
×
(
100
s
1
−
s
1
/
2
)
N/mm
2
.
Similarly, the shear flow in the segment 62 is
s
6
(τ
h
×
6
)
62
=
−
100
×
10
3
1400
×
10
4
(
−
50
)
×
6
×
d
s
1
,
0
whence
(τ
h
)
62
=
0.007143
×
(
50
s
6
)
N
/
mm
2
.
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