Civil Engineering Reference
In-Depth Information
4.18
(
5
/
16
)
=
Shear stress on base metal
=
13.4 ksi
≤
0.35(50)
≤
17.5 ksi OK
4.18
0.707
(
5
/
16
)
=
Shear stress on fillet weld throat
=
18.9 ksi
≤
0.28(70)
≤
19.6 ksi OK
F
E
=
25.1 kips
Taking moments about
F
L2
yields
4.18
(
6
)
=
T(
1.73
)
−
F
E
(
3
)
158.1
(
1.73
)
−
25.1
(
3
)
F
L1
=
=
=
33.0 kips
6
6
F
L2
=
−
F
L1
−
F
E
=
158.1
−
33.0
−
25.1
=
100.0 kips
T
Length of weld
L
1
=
33.0
/
4.18
=
7.9 in., say 8 in.
23.9 in., say 24 in.
The effect of the force eccentricity is to require that the fillet welds are
balanced such that weld
L
2 is 16 in. longer than weld
L
1. The length of weld
L
2 may be reduced by reducing the eccentricity.
If the connection length is assumed to be
(
8
Length of weld
L
2
=
100.0
/
4.18
=
+
24
)/
2
=
16 in., the shear
lag coefficient,
U
,is
U
=
(
1
−
x/L)
=
(
1
−
1.73
/
16
)
=
0.89, which is sufficiently
close to
U
=
0.90 assumed.
9.2.4.5
Eccentrically Loaded Welded Connections
Even small load eccentricities must be considered in design since welded connections
have no initial pretension (such as that achieved by the application of torque to bolts).
Many welded connections are loaded eccentrically (e.g., the connections shown in
Figures 9.4b-d)
.
Eccentric loads will result in combined shear and torsional moments
or combined shear and bending moments, depending on the direction of loading with
respect to weld orientation in the connection.
9.2.4.5.1 Connections Subjected to Shear Forces and Bending Moments
A connection such as that of Figure 9.4d is shown in greater detail in
Figure 9.5.
The
fillet welds each side of the stiffener resist both shear forces and bending moments.
The shear stress on the welds is
P
A
=
P
2
t
e
d
τ =
(9.4)
M
S
x
=
3
Pe
t
e
d
2
.
σ
b
=
(9.5)
The stress resultant is
6
e
d
2
P
2
t
e
d
2
=
τ
2
+ σ
b
=
+
f
1
.
(9.6)
