Civil Engineering Reference
In-Depth Information
As
F
D
Lk,BL
=
11.82 kN/m
<
F
LEd,1
=
84.37 kN/m, the first component from the basic
value of the bond is
q
b
L
τ
L1k
s
L0k
E
L
t
L
F
LEd
;
1
Δ
F
Lk
;
BL
F
LEd
;
1
p
100
2
?
2
:
39
?
0
:
20
?
170 000
?
1
:
4
84 370
2
Δ
F
Lk
;
BL
84 370
6
:
51
?
10
3
N
=
m
6
:
51 kN
=
m
The second component from the frictional bond is obtained with the frictional bond
stress
τ
LFk
according to DAfStb guideline part 1 annex RV K 1 as follows:
s
τ
L1k
?
s
L
0
k
t
L
?
E
L
!
!
F
LEd
;
1
b
L
?
t
L
?
E
L
2
?
t
L
?
E
L
τ
L1k
F
LEd
;
1
b
L
?
t
L
?
E
L
Δ
F
Lk
;
BF
τ
LFk
?
b
L
?
s
r
?
8
?
α
cc
?
f
cm
0
:
89
85
?
28
0
:
89
mm
2
τ
LFk
10
:
10
:
8
?
0
:
0
:
47 N
=
2
?
1
:
4
?
170 000
2
F
Lk
;
BF
0
;
47
?
100
?
200
:
96
:
39
0
@
!
1
A
2
84 370
2
:
39
?
0
:
20
84 370
100
?
1
4
?
170 000
?
1
:
100
2
4
2
?
170 000
2
:
4
?
170 000
?
1
:
92
?
10
3
N
m
The third component, caused by the curvature of the member, with
κ
k
=
24.3
10
3
and the
crack strains
ε
Lr1
and
ε
cr1
at the less heavily stressed crack edge, is
Δ
F
Lk
;
KF
s
r
?
κ
k
?
ε
Lr1
ε
cr1
h
6
;
=
m
6
:
92 kN
=
?
b
L
?
10
3
4
:
89
1
:
74
Δ
F
Lk
;
KF
201
:
96
?
24
:
3
?
10
3
?
100
?
160
20
:
24
?
10
3
N
=
m
20
:
24 kN
=
m
The admissible change in the strip force for element 3 is therefore
F
LRd
Δ
F
Lk
;
BL
Δ
F
Lk
;
BF
Δ
F
Lk
;
KF
6
:
51
6
:
92
20
:
24
Δ
22
:
45 kN
=
m
γ
BA
1
:
5
The analysis for element 3 is as follows:
Δ
F
LEd
11
:
71 kN
=
m
Δ
F
LRd
22
:
45 kN
=
m
The analysis is carried out for every concrete element between cracks. The
three components of the admissible change in the strip force depend on the acting
Search WWH ::
Custom Search