Civil Engineering Reference
In-Depth Information
Equations 9.47-9.49 can be used to eliminate
k
t
,
δθ
y
,
δθ
z
,
y
r
,
z
r
from equation
9.46, which then becomes
N
xi
=
N
x
A
i
A
i
+
M
y
A
i
z
i
−
M
z
A
i
y
i
A
i
z
i
A
i
y
i
.
(9.50)
This result demonstrates that the connector force can be obtained by super-
position of the separate components due to the centroidal force
N
x
and the prin-
cipal axis moments
M
y
,
M
z
. When there are
n
bolts of equal area, this equation
simplifies to
N
xi
=
N
x
n
+
M
y
z
i
−
M
z
y
i
z
i
y
i
.
(9.50a)
When the connectors are welds, the summations in equation 9.47 should be
replaced by integrals, so that
σ
w
=
N
x
A
+
M
y
z
−
M
z
y
I
z
(9.50b)
I
y
in which
y
,
z
are the coordinates of the weld where the stress is
σ
w
.
The assumption that only the connectors transfer compression forces through
the connection may be unrealistic when portions of the plates remain in contact.
In this case the analysis above may still be used, provided that the values
A
i
,
y
i
,
z
i
usedforthecompressionregionsrepresenttheactualcontactareasbetweenthe
plates.
9.10 Worked examples
9.10.1 Example 1 - in-plane analysis of a bolt group
Problem.
Thesemi-rigidwebsideplate(finplate)jointshowninFigure9.22aisto
transmitfactoreddesignactionsequivalenttoaverticaldownwardsforceof
Q
kN
acting at the centroid of the bolt group and a clockwise moment of 0.2
Q
kNm.
Determine the maximum bolt shear force.
Solution.
Fortheboltgroup,
(
y
i
+
z
i
)
=
4
×
(
70
2
+
105
2
)
+
4
×
(
70
2
+
35
2
)
=
88200mm
2
.
Using equation 9.40,
y
r
=
−
Q
×
10
3
×
88200
−
0.2
Q
×
10
6
×
8
=
55.1 mm
and so the most heavily loaded bolts are at the top and bottom of the right-hand
row.
For these,
r
i
=
√
{
(
55.1
+
70
)
2
+
105
2
}=
163.3mm,
and the maximum bolt force may be obtained from equation 9.3 as
F
v
,
Ed
=
−
0.2
Q
×
10
6
88200
N
=−
0.370
Q
kN.
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