Civil Engineering Reference
In-Depth Information
or
f
B
≤
1.2
F
u
≤
1.2
(
65
)
≤
78 ksi OK.
Te
ns
ile stress in the member
:
x
1.68 in.
A
e
=
=
2.85 in.
2
Tensile axial stress in the angle net section
75
2.85
=
= σ
a
=
26.3 ksi
≤
0.47(65)
≤
30.6 ksi OK
Tensile axial stress in the angle gross section
75
5.75
=
= σ
a
=
13.0 ksi
∗
0.55(50)
27.5 ksi OK
Block shear failure in angle
:
A
gt
=
∗
1.25 in.
2
2.5
(
0.5
)
=
0.75 in.
2
A
nt
=
1.25
−
0.5
(
1
)
=
10.50 in.
2
A
gv
=
2
(
9
+
1.5
)(
0.5
)
=
6.50 in.
2
A
nv
=
10.50
−
8
(
0.5
)(
1.0
)
=
P
vs
=
0.30
F
U
A
nv
+
0.50
F
U
A
nt
=
0.30
(
65
)(
6.50
)
+
0.50
(
65
)(
0.75
)
=
126.8
+
24.4
151.2 kips
Tensile ultimate strength
=
=
F
u
A
nt
=
(
65
)(
0.75
)
=
48.8 kips
253.4 kips
Therefore, tensile yielding on the gross section and shear fracture on the
net section is appropriate to consider.
P
vs
=
Shear ultimate strength
=
0.60
F
u
A
nv
=
0.60
(
65
)(
6.50
)
=
0.30
F
U
A
nv
+
0.55
F
y
A
gt
=
0.30
(
65
)(
6.50
)
+
0.55
(
50
)(
1.25
)
=
126.8
+
34.4
161.2 kips
The allowable block shear stress is 151.2 kips
=
75 kips OK
The member design is governed by the tensile fracture criterion due to
the considerable shear lag effect associated with the single angle connection
used for this secondary member.
≥
Gusset plate
Block shear failure in the gusset plate
:
A
gt
=
0.94 in.
2
2.5
(
0.375
)
=
0.56 in.
2
A
nt
=
0.94
−
0.375
(
1
)
=
9.38 in.
2
A
gv
=
2
(
9
+
3.5
)(
0.375
)
=
6.38 in.
2
A
nv
=
9.38
−
8
(
0.375
)(
1.0
)
=
P
vs
=
0.30
F
U
A
nv
+
0.50
F
U
A
nt
=
0.30
(
65
)(
6.38
)
+
0.50
(
65
)(
0.56
)
=
124.3
+
18.2
142.5 kips
Tensile ultimate strength
=
=
F
u
A
nt
=
(
65
)(
0.56
)
=
36.4 kips
248.6 kips
Therefore, tensile yielding on the gross section and shear fracture on the
net section is appropriate to consider.
P
vs
=
Shear ultimate strength
=
0.60
F
u
A
nv
=
0.60
(
65
)(
6.38
)
=
0.30
F
U
A
nv
+
0.55
F
y
A
gt
=
0.30
(
65
)(
6.38
)
+
0.55
(
50
)(
0.94
)
=
124.3
+
25.9
150.2 kips
The allowable block shear stress is 142.5 kips
=
≥
75 kips OK
Axial tension in the gusset plate
:
2
l
c
tan
(
30
◦
)
+
w
e
=
s
r
=
1.15
(
9
)
+
2.5
=
12.85 in.
75
(
12.85
)(
0.375
)
=
σ
at
=
15.6
≤
0.55
F
y
≤
27.5 ksi OK
