Civil Engineering Reference
In-Depth Information
and the torsional shear stress,
τ
T
, on the highest stressed bolts is
M
e
y
T
n
s
A
b
n
b
(x
T
+
300
(
1.25
)
τ
T
x
=
y
T
)
=
2
(
0.6
)
4
(
6
2
1.25
2
)
1.25
2
)
+
4
(
3
2
+
+
300
(
1.25
)
2
(
0.6
)(
192.5
)
=
=
1.6 ksi,
M
e
x
T
n
s
A
b
n
b
(x
T
+
300
(
6
)
2
(
0.6
)(
192.5
)
=
τ
T
y
=
y
T
)
=
7.8 ksi.
The maximum resultant shear stress in the bolts is
(
7.3
(
τ + τ
T
y
)
2
2
T
x
=
7.8
)
2
1.6
2
f
=
+ τ
+
+
=
15.2 ksi
≤
17.0 ksi OK
check bearing stress
(
15.2
)(
0.60
)
1
(
0.50
)(
1.00
)
=
σ
bc
=
18.2
≤
f
B
ksi
l
e
F
u
2
d
b
=
(
1.5
)(
65
)
2
(
7
/
8
)
f
B
=
=
55.7 ksi
assuming a minimum loaded edge distance of 1.5 in., OK
or
f
B
=
1.2
F
u
=
1.2
(
65
)
=
78 ksi.
9.3.4.3.4 Axially Loaded Connections with Bolts in Direct Tension
Connections with bolts subject to direct tension should generally be avoided in the
main members of steel railway superstructures. However, when bolts are subjected
to direct tension, the additional bolt forces created by prying action of the connection
leg must also be considered (e.g., by using Equation 9.38). The effects of the prying
action on the allowable fatigue design stresses must also be considered as shown in
Example 9.8.
Example 9.8
Design the bolted hanger-type connection shown in
Figure E9.7
for an axial
force consisting of
P
DL
=
20 kips (dead load),
P
LL+I
=
60 kips (live load plus impact).
Use 7/8 in diameter A325 bolts.
b
=
2.0
−
0.5
=
1.5 in.
a
3 in.
t
p
=
=
0.5 in.
T
3
b
3
(
1.5
)
8
(
3
)
−
t
p
20
(
0.5
)
3
20
(
20
+
60
)
T
Q
=
8
a
−
=
=
20
(
0.18
)
=
3.60 ksi
4
