Civil Engineering Reference
In-Depth Information
or
75
σ
at
=
2
(
0.375
)(
1
))(
0.375
)
=
16.5
≤
0.47
F
U
≤
30.6 ksi OK
(
12.85
−
Axial compression in the gusset plate
:
In this example
T
50 kips. However, investigate the allowable compres-
sion stresses in the gusset plate for the case of complete stress reversal
T
=
=−
C
(typical of wind bracing).
C
=−
50 kips,
f
bv
=
17 ksi (slip-resistant connection allowable bolt shear),
C
=
1.5
(
−
50
)
=−
75 kips
Allowable Compressive Strength of Member (see Chapter 6)
r
min
=
r
xy
=
1.18 in.
kL
r
min
=
0.75
(
7
)(
12
)
1.18
=
53.4
F
call
=
0.60
(
50
)
−
0.165
(
53.4
)
=
21.2 ksi
Allowable strength
121.6 kips compression
Design connection for 75 kips axial compression.
=
(
21.2
)(
5.75
)
=
Compressive stress in the gusset plate
:
C
w
e
t
p
=
75
12.85
(
0.375
)
=
σ
ac
=
15.6 ksi
0.629
E
0.65
(
6.5
)
√
12
0.375
Kl
w
r
w
=
=
39.0
≥
F
y
≥
15.2
17500
F
y
E
3
/
2
Kl
w
r
w
165.7
Kl
w
r
w
F
c
=
0.60
F
y
−
=
0.60
F
y
−
=
30,000
−
165.7
(
39.0
)
=
23,533 psi
=
23.5 ksi
≥
15.6 ksi OK
Iftheconnectionshownin
FigureE9.4
h
asacompressiondiagonalcreating
a vertical force of 35 kips and horizontal force of 25 kips in addition to the
50 kip tensile force;
P
=
T
=
50
−
25
=
25 kips
V
=
35 kips
M
=
35
(
6.5
)
=
227.5 in-kips (on section through the last line of bolts)
6.75 in.
2
A
g
=
18
(
0.375
)
=
6.00 in.
2
A
n
=
6.75
−
2
(
1
)(
0.375
)
=
0.375
(
18
)
2
/
6
19.3 in.
3
S
n
=
−
2
(
1
)(
0.375
)(
1.25
)
=
5.2 ksi (1.5V/A not used since slender beam theory not
theoretically valid)
σ
a
=
τ
v
=
35
/
6.75
=
25
/
6.00
=
4.2 ksi
σ
b
=
11.8 ksi
Use a linear interaction formula to examine combined stress effects
227.5
/
19.3
=
5.3
0.35
(
50
)
+
4.2
0.55
(
50
)
+
11.8
0.55
(
50
)
=
0.30
+
0.15
+
0.43
=
0.88
≤
1.00 OK
