Civil Engineering Reference
In-Depth Information
V
D
13 000
26
,
∗
2
v
=
=
=+
500
plfunitshear a
ttgrid line right)
2
2
R
right
lb shearatgridline3
∗
V
3
=
13 000
,
-
200 12
(
)
=
10 600
,
V
4
=
10 600
,
-
200 12
(
)
=
8200
lb
8200
26
)
∗
v
4
=
= +
315 4
.
plfunitshear at grid li
ne
4(
left
L
8200
50
v
=
= +
164
plfunitshear at grid line ight
4(
)
4
R
The shear at grid line 5 is equal to +104 plf.
Find the chord forces (without opening) at grid line 3.
wx
2
200 27
2
()
2
MRx
=- =
16 000 27
,
(
)
-
=
359 100
,
ft-lb
3
1
2
M
D
359 100
50
,
F
==
3
=
7182
lb at Aand D
3
Resolve shears and chord forces on elements I through IV (Fig. 5.9). Start at grid line
4 to sum shears.
Shear distribution to elements I and III:
V
4
=
8200
lb previously calculated.
The shears at element I and III are distributed to each element in accordance with
their depth.
DV
D
10 8200
26
(
)
16 8200
26
(
)
V
=
AB
-
4
=
=
3153 8
.
lb
V
=
=
5046 2
.
lb
∑
4
AB
4
CD
left
8200
26
v
== =
v
315 4
.
plfunitshear at grid
line nelementsIand III
4
AB
CD
Element I:
Shear at grid line 3:
4629 8
10
.
V
=
3153 8
.
+
123 12
(
)
=
4629 8
.
lb
v
=
=+
4
63 plf
3
3
AB
F
3
=
7182
lb compression as previously calcu
lated
A
F
3
=
0
as previously assumed
B
∗
Note:
The actual shear applied to the segments between grid lines 2 and 4 is not uniform (i.e.,
VD
2
/
AB C+
).
The actual shears vary in each segment and will be determined by analyzing each segment separately.
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