Civil Engineering Reference
In-Depth Information
5600
2
V
3
=
=
2800
lb load perwallpanel at
3
rd floor
5600
2
=
=
2800
lb load perwallpanel at
2
nd floor
V
2
The typical distances to the forces from the point of summing moments are shown
at the lower left of the figure.
Third floor:
V
L
1400
8
r
v
=
=
=
175
plfunitwallshear a
tt dfloor
3
wall
SW
1
7
T
3
=
[
1400 10
(
)
-
06800 3 875
. (
)(.
)
-
06250 7 875
. (
)(.
)]
=
1461 2
.
lb tension
5
1
75
C
3
=
[
1400 10
(
)(
+
800 3 625
)( .
)
+
(
650 7 625
)(.
)
-
250 037
( .
5
)]
.
=
2901 7
.
lb compression
Second floor:
VV
L
+
1400
+
2800
r
3
525
shearatndfloor
2
v
=
=
=
plfunitwall
wall
8
SW
1
75
T
2
=
[
1400 20
(
)
+
2800 10
(
)
-
06800 23875
. (
)()(.
)
-
0625
. (
0
+
1000 7 875
)( .
)]
.
=
6183 2
. btension
C
2
=
[
1400 20
(
)
+
2800 10
(
)(
+
800 23625
)( )( .
)
+
(
650
+
1200
)( .
7 625
)
1
75
-
(
250
+
1000 0 375
)(.
)]
=
10 058 3
,
.
lb compression
.
Compression perpendicular to the grain should be checked for two studs bearing
on the wall bottom plate.
First floor:
VVV
L
++
=
1400
+
2800
+
2800
r
3
2
875
nitshear at
1
st floor
v
=
=
plfu
wall
8
SW
T
1
=
[
1400 32
(
)
+
2800 22
(
+-
12
)
061600
. (
+
960 3 875
)(.
)
1
75
-
06250
.(
+
2000 7 875
)( .
)]
=
16 445 6lbtension
,
.
.
C
1
=
[
1400 32
(
)
+
2800 22
(
++ +
12
)(
1600
960 3 625
)( .
)
+
(
650
+
2400 7 625
)( .
)
1
75
-
(
250
+
2000 0 375
)(.
)]
=
22 892 3
,
.
lb compression
.
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