Civil Engineering Reference
In-Depth Information
f
ck
200
λ
150
β
0
:
35
30
200
23
9
150
:
β
0
:
35
0
:
34
8
f
cm
N
=
mm
2
16
:
?
β
0
;
k
M
0
;
Eqp
M
Ed
φ
ef
k
?
q
3
p
?
1
:
0
20
:
7
16
:
8
φ
ef
0
:
39
?
48
:
5
0
:
45
Using these variables it is possible to calculate the maximum acting moment:
l
2
π
M
Ed
N
Rd
?
e
tot
?
ξ
1
?
ξ
2
?
ϕ
bal
?
K
ϕ
2
3000
2
π
6642
:
4
?
10
3
?
0
:
58
?
1
?
1
:
63
?
10
5
?
10
6
7
:
5
?
1
:
15
115
:
7 kNm
2
It is now necessary to calculate the resistance of the cross-section with the second part of
Eq. (RV 6.63) according to the DAfStb guideline [1, 2]:
sin
3
1
γ
LG
?
2
3
?
α
1
?
f
cck
?
A
c
?
D
2
?
π
?
θ
M
Rd
π
1
γ
s
?
f
syk
?
A
s
?
D
2
?
sin
π
?
θ
sin
π
?
θ
c
t
π
sin
3
1
2
3
?
0
500
2
?
π
?
0
:
809
95
?
1964
?
10
2
?
10
6
M
Rd
:
953
?
32
:
35
?
?
1
:
π
1
1
:
15
?
500
?
5890
?
500
2
?
sin
π
?
0
:
886
sin
π
?
0
?
10
6
115
:
1 kNm
π
As the resistance is equal to the action, the relative angle
θ
chosen was correct.
M
Rd
115
:
7 kNm
M
Ed
115
:
1 kNm
As the acting axial load is less than the axial load resistance of the column, the column
load-carrying capacity is satisfactory.
N
Rd
6642
:
4kN
N
Ed
6469
:
8kN
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