Civil Engineering Reference
In-Depth Information
Strut forces along grid line A: The direction of transfer shears determines if the
member is in tension or compression. Starting at grid line 1,
At grid line 2,
=
+
333 3
.
+
111 1
.
40
8890
F
2
()
=
lb compression
A
2
At grid line 3,
85 9
.
+
145 9
2
.
F
3
=
8890
-
()
15
=
7151 5
.
lb compress
ion
A
At diaph. C.l,
020
2
+
F
CL
=
7151 5
.
+
( )
5
=
7201 5
.
lb compression
At grid line 4,
060
2
+
F
4
=
7201 5
.
-
(
15
)
=
6751 5
.
lb compression
At grid line 5,
420
+
480
F
5
=
6751 5
.
-
(
15
)
=
15
.
lb compression
2
Close enough, the diagram closes.
Strut force along grid line B, starting at grid line 4: The chord force on the right side
of grid line 5 was calculated as 13,500 lb compression. The force on the left side of grid
line 5 is equal to
(
480
+
420
)
+
(
420
+
480
)
()
F
5
=
15
=
13500
lb compr
ession
B
2
Also note that
600
+
300
F
56
-
=
()
30
=
13500
lb compression
2
Strut force along grid line C, starting at grid line 1: The chord force on the left side
of grid line 2 was calculated as 8890 lb tension. The force on the right side of line 2 is
equal to
(
446 7
.
+
145 9
.) (
+
506 7859
.
+
.)
()
F
2
=
15
=
8890
lb
tension
C
2
Therefore the diagran closes.
Strut force along grid line D, starting at grid line 2:
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