Civil Engineering Reference
In-Depth Information
The net shear at the shear wall is equal to the wall shear minus the diaphragm shear.
v
SW
= 161 3
.
plf
v
=- =
v
v
161 3451
.
-
.
=+
116 2
.
plf
net
SW
iaph
Since the wall shears are acting in an opposite direction to the strut force, they must be
add to the negative value of the strut force. The force at the right end of SW3 is equal to
F
=-
1030
+
116 28
.()
= -
100 5
.
lb compression
The force at the left end of SW4 is equal to
F
=-
100 545145
.
-
. ()
= -
2129 9
. bcompression
The net shear at SW4 is also 116.2 plf, acting opposite to the strut force; therefore the
force at grid line 4 is equal to
F
=-
. b
2129 9
.
+
116 215
. ()
= -
386 9
Notice that the force diagram does not close to zero; therefore, an error exists. This
reinforces the need to confirm that all force diagrams close before verifying that a complete,
workable design exists. However, before abandoning the analysis, determination of the
strut forces along grid line A should be completed. The additional information gained
will make it easier to determine what caused the error. Starting at grid line 1, the force
at grid line 2A is equal to
() lb compression
The force at the left end of SW2 is equal to
F
=-
28 25
=-
700
F
=-
700
-
32 05 5
.
()
= -
860
lbcompression
The net shear at the shear wall is equal to
R
L
5000
10
v
=
A
SW
=
=
500
plf
SW
2
v
=- =- =
v
v
500
32 05
.
467 95
.
plf
net
SW
iaph
Acting in theopposite directiontothe strut force. The force at the right end of
SW2 is equal to
F
=-
860
+
467 95 10
.
()
= +
3819 3
.
lb tension
The force at grid line 3 is equal to
F
=
lb tension
3819 332055
.
-
.
()
=
3659
The force at grid line 4 is equal to
F
=
lbtension
3659
-
40 980
.( )
=+
387
This force diagram also does not close to zero by the same amount. Figure 3.28 shows
how to correct the problem. At grid line C, the shear wall shears are not high enough to
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