Civil Engineering Reference
In-Depth Information
Completing the remaining shear forces on the sheathing element shows that the shear
is negative.
2256
15
V
=
1875 2
.
-
4131 2
.
=-
2256
lb
v
=
=
150 4
.plf
BC
BC
Completing the remaining shear forces on the sheathing element shows that the shear
is positive.
Net diaphragm shears:
Location (Grid Line)
Net Shear
4 from A to B
v
= +164 -(160.9) = +3.1 plf
4 from B to C
v
= +164 +(150.4) = +314.4 plf
4 from C to D
v
= +164 - (125) = +39 plf
5 from A to B (left side)
v
= +104 - (160.9) = -56.9 plf
5 from B to C (left side)
v
= +104 + (150.4) = +254.4 plf
5 from C to D (left side)
v
= +104 - (125) = -21 plf
Collector forces at grid line 4, summing from grid line D:
Location (Grid Line) Force
4 at C
F
2 C
= (315.4 -39)16 = 4422.4 lb compression
4 at B
F
2B
= 4422.4 - (314.4)24 = -3123.2 lb tension
4 at A
F
2A
= -3123.2 + (315.4 -3.1)10 = 0 lb tension
Collector forces at grid line 5, summing from grid line D:
Location (Grid Line)
Force
5 at C
F
2C
= (104 + 21)16 = 2000 lb tension
5 at B
F
2B
= 2000 - (254.4 - 104)24 = 1609.6 lb compression
5 at A
F
2A
= 1609.6 - (104 + 56.9)10 = 0 lb compression
Determine the longitudinal chord/collector forces (Fig. 5.12):
Collector forces at grid line B, summing from grid line 1:
Location (Grid Line)
Force
(
508 8794
.
-
.) (
+
448 8194
.
-
.)
()
Bat
2
F
=
15
B
2
=
6441
lb compression
610
.
6
+
463
Bat
3
F
=+
6441
-
()
12
3
B
2
=
0
lb compression
463
+
315 4
2
.
Bat
4
F
=
0
-
()
12
4
B
=-
4670 4
. btension
(
315 431
.
-+ +
.) (
254 4569
.
.)
(15
B
at
5
F
=-
4670 4
.
+
)
B
2
= lb tension
0
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