Civil Engineering Reference
In-Depth Information
404 in.
4
I
x
=
53.8 in.
3
S
x
=
r
x
=
5.24 in.
11.0 in.
4
I
y
=
3.78 in.
3
S
y
=
r
y
=
0.867 in.
x
=
0.80 in.
For laced member:
29.4 in.
2
A
=
2
(
14.7
)
=
808 in.
4
I
x
=
2
(
404
)
=
0.80
)
2
504 in.
4
I
y
=
2
(
11.0
)
+
2
(
14.7
)(
3.25
+
=
504
29.4
=
r
y
=
4.14 in.
L
r
min
=
(
19.55
)(
12
)
4.14
=
56.7
≤
100 OK
5.034
E
KL
r
min
=
0.75
(
19.55
)(
12
)
4.14
=
42.5
≤
F
y
≤
121
0.629
E
KL
r
min
=
42.5
≥
F
y
≥
15.
Therefore,
17,500
F
y
E
3
/
2
KL
r
min
0.166
KL
r
min
F
all
=
0.60
F
y
−
=
30
−
=
30
−
7.1
=
22.9 ksi
P
all
=
F
all
(A)
=
22.9
(
29.4
)
=
675
≥
550 kips OK.
If the effects of shear deformation are included (Equations 6.43 and 6.48b),
1
(L/r)
2
13.2
A
A
plb
sin
α =
+
cos
2
φ
φ
1
13.2
(
56.7
)
2
29.4
2
(
4
)(
0.5
)
sin 30.6
◦
cos
2
30.6
◦
=
+
=
1.04
P
all
α
675
1.08
=
P
all
=
=
625
≥
550 kips OK.
2
