Civil Engineering Reference
In-Depth Information
Solution.
F
p
,
C
=
0.7
×
800
×
245N
=
137.2kN
EC3-1-8 3.9.1(2)
k
s
=
1.0
EC3-1-8T3.6
µ
=
0.4
EC3-1-8T3.7
γ
M
3,
ser
=
1.1
EC3-1-8 2.2(2)
F
s
,
Rd
=
1.0
×
0.4
×
137.2
/
1.1
=
49.9kN
EC3-1-8 3.9.1(1)
and using the solution of example 1,
Q
sL
=
49.9
/
0.370
=
134.7.
9.10.7 Example 7 - out-of-plane resistance of a bolt group
Problem.
The flexible end plate joint shown in Figure 9.22b is to transmit fac-
tored design actions equivalent to a downwards force of
Q
kN acting at the
centroid of the bolt group and an out-of-plane moment of
M
x
=−
0.05
Q
kNm. Determine the maximum bolt forces, and the design resistance of the bolt
group.
Bolt group analysis.
z
i
=
4
×
105
2
+
4
×
35
2
mm
2
=
49000mm
2
,
and using equation 9.50a, the maximum bolt tension is
F
t
,
Ed
=
(
−
0.05
Q
×
10
6
)
×
(
−
35
−
70
)
49 000
N
=
0.107
Q
kN.
(Aless-conservative value of
F
t
,
Ed
=
0.0824
Q
kN is obtained if the centre of
compression is assumed to be at the lowest pair of bolts so that the lever arm to
the highest pair of bolts is 175mm).
The average bolt shear is
F
v
,
Ed
=
(
Q
×
10
3
)
/8N
=
0.125
Q
kN.
Plastic resistance of plate
.
If the web thickness is 8.8 mm,
m
=
90
/
2
−
8.8
/
2
−
0.8
×
6
=
35.8mm EC3-1-8 F6.8
l
eff
=
2
×
35.8
+
0.625
×
30
+
35
=
125.4mm EC3-1-8T6.4
M
pl
,1,
Rd
=
0.25
×
125.4
×
8
2
×
355
/
1.0Nmm
=
0.712kNm EC3-1-8T6.2
F
T
,1,
Rd
=
4
×
0.712
×
10
6
/
30N
=
94.9kN
EC3-1-8T6.2
and so
F
T
,1,
Rd
≥
2
×
0.107
Q
, whence
Q
p
≤
443.0kN.
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